Definite Integrals – JEE Main MathematicsPractice Questions & PYQs

The integral $11\alpha(10,6) = 11\int_0^2 t^{10}(1+3t)^6 dt$ is best handled by applying integration by parts with $u = (1+3t)^6$ and $dv = t^{10} dt$, which leads to $du = 18(1+3t)^5 dt$ and $v = \frac{t^{11}}{11}$. Substituting these into the integration by parts formula and multiplying by 11 yields 11\alpha(10,6) = \[t^{11}(1+3t)^6]_0^2 - 18\alpha(11,5)$. Adding$18\alpha(11,5)$to both sides isolates the expression$11\alpha(10,6) + 18\alpha(11,5) = $[t^{11}(1+3t)^6]_0^2$, which evaluates to$2^{11}(7)^6$. Because$2^{11}(7)^6$can be rewritten as$32(14)^6$, the value of$p$ is 32.

The presence of an absolute value within an integral function implies that the integrand's behavior changes at the point where the expression inside vanishes, requiring us to evaluate the integral across distinct intervals to identify where the minimum occurs. For $x \leq 0$, the function strictly decreases, and for $x \geq 2$, it strictly increases, effectively localizing the minimum value to the interior range $(0, 2)$. In this interval, the integral $f(x) = \int_{0}^{2} e^{|x-t|} dt$ is decomposed into $\int_{0}^{x} e^{x-t} dt + \int_{x}^{2} e^{t-x} dt$, which evaluates to $(e^x - 1) + (e^{2-x} - 1)$, resulting in $f(x) = e^x + e^{2-x} - 2$.

Since we aim to minimize $f(x)$ for $x \in (0, 2)$, we can apply the arithmetic mean-geometric mean inequality to the variable terms $e^x$ and $e^{2-x}$. Because their product $e^x \cdot e^{2-x}$ equals $e^2$, the inequality $\frac{e^x + e^{2-x}}{2} \geq \sqrt{e^x \cdot e^{2-x}}$ simplifies to $e^x + e^{2-x} \geq 2e$. Subtracting 2 from both sides of this expression confirms that the minimum value of the function is $2e - 2$, which can be written as $2(e-1)$.

The evaluation of this integral relies on factoring the cubic denominator into linear terms of $e^x$, which allows the integrand to be simplified using partial fraction decomposition. The expression $e^{3x}+6e^{2x}+11e^{x}+6$ factors into $(e^x+1)(e^x+2)(e^x+3)$, and the integrand can be rewritten as $3 \left( \frac{1}{e^x+1} - \frac{2}{e^x+2} + \frac{1}{e^x+3} \right)$. Multiplying the numerator and denominator of each fraction by $e^{-x}$ transforms these into terms of the form $\frac{e^{-x}}{1+ae^{-x}}$, which are easily integrated to logarithmic forms.

Integrating term by term produces $-3 \ln(1+e^{-x}) + 3 \ln(1+2e^{-x}) - \ln(1+3e^{-x})$, which must be evaluated over the interval from $0$ to $\infty$. As $x$ approaches $\infty$, all terms involving $e^{-x}$ vanish to $\ln(1)=0$, while at the lower limit of $x=0$, the expression simplifies to $-3\ln(2) + 3\ln(3) - \ln(4)$. Subtracting the value at the lower limit from the value at the upper limit results in $3\ln(2) - 3\ln(3) + \ln(4)$, which simplifies to $\log_e \left( \frac{32}{27} \right)$.

When evaluating this limit, substituting $x=0$ results in the indeterminate form $\frac{0}{0}$ because the definite integral over a zero-width interval is zero and the denominator $x^4$ also vanishes. L'Hospital's Rule simplifies this calculation by requiring the differentiation of both the numerator and denominator with respect to $x$. By applying the Fundamental Theorem of Calculus to the integral $\int_{0}^{x} \frac{t^3}{t^{6+}1} dt$, its derivative with respect to $x$ is simply $\frac{x^3}{x^{6+}1}$, while the derivative of the denominator $x^4$ is $4x^3$. The expression then becomes:
$\lim_{x \to 0} \frac{48 \cdot \frac{x^3}{x^{6+}1}}{4x^3}$
Canceling the common factor of $x^3$ in the numerator and denominator reduces the expression to $\frac{12}{x^{6+}1}$. As $x$ approaches zero, this value converges to $\frac{12}{0^{6+}1}$, which equals 12.

When dealing with definite integrals featuring reciprocal limits, the substitution $x = \frac{1}{t}$ is highly effective at simplifying the integrand. Applying this transformation to the integral $I = \int_{1/2}^2 \frac{\tan^{-1} x}{x} dx$ changes the limits from $2$ to $1/2$ and incorporates the differential $dx = -\frac{1}{t^2} dt$, which ultimately transforms the expression into $I = \int_{1/2}^2 \frac{\tan^{-1}(1/x)}{x} dx$ after reverting the dummy variable back to $x$.

Using the trigonometric identity $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$, we can sum the original integral and its transformed version to obtain $2I = \int_{1/2}^2 \frac{\tan^{-1} x + \cot^{-1} x}{x} dx$. This substitution simplifies the integrand significantly to $2I = \frac{\pi}{2} \int_{1/2}^2 \frac{1}{x} dx$. Evaluating this logarithmic integral results in $2I = \frac{\pi}{2} (\ln 2 - \ln \frac{1}{2})$, which simplifies to $\frac{\pi}{2} (2 \ln 2)$ and yields the final value of $\frac{\pi}{2} \log_e 2$.

Because the function $|100x^2 - 1|$ is even, the integral over the symmetric range from $-0.15$ to $0.15$ is equivalent to twice the integral from $0$ to $0.15$. The expression $100x^2 - 1$ equals zero at $x = 0.1$, which marks the point where the function changes its sign. Therefore, the integral must be partitioned into two distinct intervals, $(0, 0.1)$ and $(0.1, 0.15)$, to remove the absolute value signs correctly.

On the interval $(0, 0.1)$, the term $100x^2 - 1$ is negative, making $|100x^2 - 1|$ equal to $1 - 100x^2$. On the interval $(0.1, 0.15)$, the term $100x^2 - 1$ is positive, so the expression remains $100x^2 - 1$. The integral expression becomes $2 \left( \int_{0}^{0.1} (1 - 100x^2) dx + \int_{0.1}^{0.15} (100x^2 - 1) dx \right)$. Evaluating these integrals, we obtain 2 \left([x - \frac{100}{3}x^3]_0^{0.1} + \[\frac{100}{3}x^3 - x]_{0.1}^{0.15} \right)$. Computing the definite integrals for each part and combining the results yields the final value$\frac{575}{3000}$, which confirms that$k$is$575$.

The greatest integer function $[\csc x]$ assumes the value $1$ for the interval $x \in [\pi/6, 5\pi/6]$, because $\csc x$ remains between $1$ and $2$ throughout this range. Consequently, the first integral $\int_{\pi/6}^{5\pi/6} 8[\csc x] \, dx$ simplifies to $8(\frac{5\pi}{6} - \frac{\pi}{6}) = \frac{16\pi}{3}$.

For the second component, involving the integral of $[\cot x]$, the property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$ reveals that $I = \int_{\pi/6}^{5\pi/6} [\cot x] \, dx$ is equal to $\int_{\pi/6}^{5\pi/6} [\cot(\pi-x)] \, dx$, which is $\int_{\pi/6}^{5\pi/6} [-\cot x] \, dx$. Adding these two expressions yields $2I = \int_{\pi/6}^{5\pi/6} ([\cot x] + [-\cot x]) \, dx$. Since $[\theta] + [-\theta] = -1$ for non-integer $\theta$, the integrand reduces to $-1$, leading to $2I = -(\frac{5\pi}{6} - \frac{\pi}{6}) = -\frac{2\pi}{3}$, which means $I = -\frac{\pi}{3}$. Substituting these results into the initial expression gives $\frac{2}{\pi} (\frac{16\pi}{3} - 5(-\frac{\pi}{3})) = \frac{2}{\pi} (\frac{21\pi}{3}) = 14$.

The substitution $t = \sin x$ with $dt = \cos x \, dx$ transforms the integral into the form $f_n = \int_0^1 (\sum_{k=1}^n t^{k-1})(\sum_{k=1}^n (2k-1)t^{k-1}) \, dt$. This simplifies the problem into the product of two finite polynomial sums, where $P_n(t) = \sum_{k=1}^n t^{k-1}$ and $Q_n(t) = \sum_{k=1}^n (2k-1)t^{k-1}$, so $f_n = \int_0^1 P_n(t) Q_n(t) \, dt$. As $n$ increments from $20$ to $21$, the sums expand by additional terms, specifically $P_{21}(t) = P_{20}(t) + t^{20}$ and $Q_{21}(t) = Q_{20}(t) + 41t^{20}$. The difference $f_{21} - f_{20}$ is equivalent to the integral $\int_0^1 [P_{21}(t) Q_{21}(t) - P_{20}(t) Q_{20}(t)] \, dt$, which simplifies to \int_0^1 \[P_{21}(t) (41t^{20}) + Q_{20}(t) t^{20}]$ , dt$after expanding the product and accounting for the shared$P_{20}(t) Q_{20}(t)$ term.

Evaluating this integral involves combining the terms of matching powers. For powers $t^{20}$ through $t^{39}$, the coefficient of each term $t^{m}$ is the sum of the corresponding components from $Q_{20}(t) t^{20}$ and $P_{21}(t) (41t^{20})$, resulting in values like $(1+41)$ for $t^{20}$, $(3+41)$ for $t^{21}$, and $(39+41)$ for $t^{39}$. Integrating these terms over $[0, 1]$ yields ratios such as $42/21, 44/22, \dots, 80/40$, each of which equals $2$. There are exactly $20$ such terms from $t^{20}$ to $t^{39}$, contributing $20 \times 2 = 40$ to the total. The remaining term from the expansion is $41t^{40}$, which integrates to $41/41 = 1$. Adding these results together, the final value is $40 + 1 = 41$.

The underlying principle is using substitution to transform an integral where the factor outside the parenthetical expression represents the derivative of the expression inside. Setting $t = 2x^{21}+3x^{14}+6x^7$ leads to the differential $dt = (42x^{20}+42x^{13}+42x^6)dx$, which factors neatly to $dt = 42(x^{20}+x^{13}+x^6)dx$. This allows replacing $(x^{20}+x^{13}+x^6)dx$ with $\frac{1}{42}dt$. Updating the integration limits from $x=0$ to $t=0$ and $x=1$ to $t=11$, the integral transforms into $\frac{1}{42} \int_{0}^{11} t^{1/7} dt$. Evaluating this yields $\frac{1}{42} \left[ \frac{7}{8} t^{8/7} \right]_{0}^{11}$, which simplifies to $\frac{1}{48} (11)^{8/7}$. Comparing this to $\frac{1}{l}(11)^{m/n}$ gives $l=48$, $m=8$, and $n=7$, and summing these values results in $l+m+n = 48+8+7 = 63$.

When integrating an absolute value function like $|x^{2-}3x+2|$, we must first determine the points where the expression inside the modulus changes sign. By factoring the quadratic into $(x-1)(x-2)$, we see that the expression is positive on the intervals $(0, 1)$ and $(2, 3)$, but negative on the interval $(1, 2)$. To evaluate the integral correctly, we split the domain at the roots $x=1$ and $x=2$, allowing us to remove the modulus bars by reversing the sign on the interval where the function is negative:
$\int_{0}^{3} |x^{2-}3x+2| dx = \int_{0}^{1} (x^{2-}3x+2) dx - \int_{1}^{2} (x^{2-}3x+2) dx + \int_{2}^{3} (x^{2-}3x+2) dx$
Integrating the polynomial $x^{2-}3x+2$ yields the antiderivative $\frac{x^3}{3} - \frac{3x^2}{2} + 2x$. Applying the limits of integration, the first part over the interval $(0, 1)$ results in $\frac{5}{6}$. The middle portion over $(1, 2)$ evaluates to $-\frac{1}{6}$, but because we flipped the sign of this region, it contributes $+\frac{1}{6}$ to the total area. The final segment over $(2, 3)$ evaluates to $\frac{5}{6}$ again. Summing these values gives $\frac{5}{6} + \frac{1}{6} + \frac{5}{6} = \frac{11}{6}$, which we then multiply by 12 to obtain 22.

The limit of a series where the number of terms grows with $n$ can be evaluated by expressing it as a Riemann integral. By rewriting the summation $\frac{1}{1+n} + \frac{1}{2+n} + \dots + \frac{1}{2n}$ as $\sum_{r=1}^{n} \frac{1}{n+r}$, we prepare the expression for conversion. Factoring $n$ from the denominator inside the summation yields $\sum_{r=1}^{n} \frac{1}{n(1 + \frac{r}{n})}$, which effectively separates the components into a $\frac{1}{n}$ term and a function of $\frac{r}{n}$.

As $n$ tends toward infinity, the definition of a Riemann sum allows us to replace the discrete summation with a definite integral. The term $\frac{r}{n}$ behaves as the continuous variable $x$, while the boundaries of the sum, as $n$ becomes arbitrarily large, correspond to the interval from $0$ to $1$. Consequently, the expression simplifies to the definite integral $\int_{0}^{1} \frac{1}{1+x} \, dx$.

Evaluating this integral results in the natural logarithm function $\log_{e}(1+x)$, which must be calculated across the boundaries of $0$ and $1$. Applying the fundamental theorem of calculus provides $\log_{e}(1+1) - \log_{e}(1+0)$, which simplifies to $\log_{e}(2) - \log_{e}(1)$. Since the logarithm of $1$ is $0$, the final result is $\log_{e} 2$.

The integral $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{2+3\sin x}{\sin x(1+\cos x)} dx$ is best addressed by decomposing the integrand into two simpler terms using the linearity property of integration, which allows us to evaluate $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{2}{\sin x(1+\cos x)} dx$ and $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{3}{1+\cos x} dx$ independently.

Focusing first on $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{3}{1+\cos x} dx$, we apply the trigonometric half-angle identity $1+\cos x = 2\cos^2(\frac{x}{2})$, which transforms the integrand into $\frac{3}{2}\sec^2(\frac{x}{2})$. The antiderivative of $\frac{3}{2}\sec^2(\frac{x}{2})$ is $3\tan(\frac{x}{2})$, and evaluating this from $\frac{\pi}{3}$ to $\frac{\pi}{2}$ gives $3(\tan(\frac{\pi}{4}) - \tan(\frac{\pi}{6}))$. Substituting the trigonometric values $3(1 - \frac{1}{\sqrt{3}})$ simplifies to $3 - \sqrt{3}$.

To solve $\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{2}{\sin x(1+\cos x)} dx$, we use the Weierstrass substitution $t = \tan(\frac{x}{2})$, where $dx = \frac{2dt}{1+t^2}$, $\sin x = \frac{2t}{1+t^2}$, and $\cos x = \frac{1-t^2}{1+t^2}$. These substitutions simplify the integrand to $\frac{1+t^2}{t}$, and changing the limits for $x = \frac{\pi}{3}$ to $x = \frac{\pi}{2}$ gives the range $t = \frac{1}{\sqrt{3}}$ to $t = 1$. The integral becomes $\int_{\frac{1}{\sqrt{3}}}^{1} (\frac{1}{t} + t) dt$, which evaluates to $[\ln|t| + \frac{t^2}{2}]_{\frac{1}{\sqrt{3}}}^{1}$. Computing this results in $(\ln 1 + \frac{1}{2}) - (\ln \frac{1}{\sqrt{3}} + \frac{1}{6})$, which equals $\frac{1}{2} - (\frac{1}{6} - \log_e \sqrt{3})$, simplifying to $\frac{1}{3} + \log_e \sqrt{3}$.

Combining the two evaluated parts yields $3 - \sqrt{3} + \frac{1}{3} + \log_e \sqrt{3}$, which simplifies to the final result of $\frac{10}{3} - \sqrt{3} + \log_e \sqrt{3}$.

The core principle for evaluating this integral is the property $\int_{0}^{\pi} f(x) \, dx = \int_{0}^{\pi} f(\pi - x) \, dx$. Because the cosine function satisfies $\cos(\pi - x) = -\cos x$, applying this symmetry allows the exponential terms in the numerator and denominator to simplify effectively, leaving the integral $2I = \int_{0}^{\pi} (1 + \cos x \cos 3x + \cos^2 x + \cos^3 x \cos 3x) \, dx$.

Evaluating each term over the interval $[0, \pi]$ yields distinct results. The integral of the constant $1$ results in $\pi$, while $\cos x \cos 3x$ integrates to zero due to the orthogonality of trigonometric functions on this interval. The term $\cos^2 x$ contributes $\frac{\pi}{2}$, and expressing $\cos^3 x$ in terms of multiple angles allows the term $\cos^3 x \cos 3x$ to be calculated as $\frac{\pi}{8}$. Aggregating these values gives $2I = \pi + \frac{\pi}{2} + \frac{\pi}{8} = \frac{13\pi}{8}$. Consequently, $I = \frac{13\pi}{16}$, which confirms that $k = 13$.

To integrate the greatest integer function $[x^2]$, we must partition the integration range into smaller sub-intervals where the integrand remains constant. Since the function changes its value whenever $x^2$ reaches an integer, we split the domain from $0$ to $2.4$ based on the points where $x$ corresponds to integer squares, specifically $0, 1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}$, and $2.4$. On each sub-interval, the function behaves as a constant, allowing the integral to be computed as that constant multiplied by the length of the interval.

Evaluating these segments, the integral from $0$ to $1$ is $0$, while from $1$ to $\sqrt{2}$ it equals $1(\sqrt{2} - 1)$, which simplifies to $\sqrt{2} - 1$. For the interval from $\sqrt{2}$ to $\sqrt{3}$, the value is $2(\sqrt{3} - \sqrt{2}) = 2\sqrt{3} - 2\sqrt{2}$, and from $\sqrt{3}$ to $2$ it is $3(2 - \sqrt{3}) = 6 - 3\sqrt{3}$. Continuing, the integral from $2$ to $\sqrt{5}$ becomes $4(\sqrt{5} - 2) = 4\sqrt{5} - 8$, and the final segment from $\sqrt{5}$ to $2.4$ yields $5(2.4 - \sqrt{5}) = 12 - 5\sqrt{5}$. Combining these results, the constants are $-1 + 6 - 8 + 12 = 9$, while the coefficients for the square root terms are $1 - 2 = -1$ for $\sqrt{2}$, $2 - 3 = -1$ for $\sqrt{3}$, and $4 - 5 = -1$ for $\sqrt{5}$. The final expression is $9 - 1\sqrt{2} - 1\sqrt{3} - 1\sqrt{5}$, meaning the sum of the coefficients $\alpha + \beta + \gamma + \delta$ is $9 - 1 - 1 - 1$, which equals $6$.

The integral relies on King's Property, $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$, which is specifically designed for evaluating trigonometric functions over the interval from $0$ to $\frac{\pi}{2}$. Swapping sine and cosine terms through this substitution allows the integrand to simplify significantly.

Applying this property to $I = \frac{8}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}} dx$ transforms the numerator into $(\sin x)^{2023}$ while leaving the denominator unchanged, resulting in the expression $I = \frac{8}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{(\sin x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}} dx$. Summing these two versions of the integral yields $2I = \frac{8}{\pi} \int_{0}^{\frac{\pi}{2}} 1 dx$, which simplifies to $2I = \frac{8}{\pi} \cdot \frac{\pi}{2} = 4$. Solving for $I$ leads to the final value of $2$.

The integral is evaluated using the standard inverse trigonometric identity $\int \frac{dt}{\sqrt{a^{2-}t^2}} = \sin^{-1}\left(\frac{t}{a}\right) + C$. Factoring $2$ out of the square root in the denominator transforms $\sqrt{9-4x^2}$ into $2\sqrt{(\frac{3}{2})^{2-}x^2}$. This allows us to rewrite the expression and evaluate the integral as follows:

$\int_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}} \frac{24}{\sqrt{(\frac{3}{2})^{2-}x^2}} dx = 24 \left[ \sin^{-1}\left(\frac{2x}{3}\right) \right]_{\frac{3\sqrt{2}}{4}}^{\frac{3\sqrt{3}}{4}}$

Evaluating the antiderivative at the provided upper and lower limits leads to the expression:

$24 \left( \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) - \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \right)$

Since the inverse sine values correspond to $\frac{\pi}{3}$ and $\frac{\pi}{4}$ respectively, the calculation becomes $24 \left( \frac{\pi}{3} - \frac{\pi}{4} \right)$, which simplifies to $24 \cdot \frac{\pi}{12} = 2\pi$.

By applying the definite integral property $\int_{-a}^{a} f(x) \, dx = \int_{-a}^{a} f(-x) \, dx$ to replace $x$ with $-x$ in $I = \int_{-\pi/4}^{\pi/4} \frac{x+\pi/4}{2-\cos 2x} \, dx$, we obtain the equivalent expression $I = \int_{-\pi/4}^{\pi/4} \frac{-x+\pi/4}{2-\cos 2x} \, dx$. Adding these two versions of the integral yields $2I = \int_{-\pi/4}^{\pi/4} \frac{\pi/2}{2-\cos 2x} \, dx$, which simplifies to $I = \frac{\pi}{2} \int_{0}^{\pi/4} \frac{1}{2-\cos 2x} \, dx$ due to the even nature of the denominator. Substituting $\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$ and using the variable change $t = \tan x$ converts the integral into $\frac{\pi}{2} \int_{0}^{1} \frac{dt}{1+3t^2}$. Evaluating this expression results in \frac{\pi}{2\sqrt{3}} \[\tan^{-1}(\sqrt{3}t)]_{0}^{1}$, leading to the final value of$\frac{\pi^2}{6\sqrt{3}}$.

The presence of an absolute value within a definite integral requires splitting the integration range at the point where the function inside changes sign. Since the natural logarithm $\ln x$ is negative when $x < 1$ and positive when $x > 1$, the integral over the interval $[\frac{1}{3}, 3]$ is split at $x = 1$ into two separate definite integrals: one from $\frac{1}{3}$ to $1$ involving $-\ln x$, and one from $1$ to $3$ involving $\ln x$.

Calculating the antiderivative of $\ln x$ yields $x \ln x - x$. Applying the boundaries to the first segment gives $-\left[ x \ln x - x \right]_{\frac{1}{3}}^{1}$, which simplifies to $-\left( -1 - (\frac{1}{3} \ln \frac{1}{3} - \frac{1}{3}) \right)$, resulting in $\frac{2}{3} - \frac{1}{3} \ln 3$. For the second segment, evaluating $\left[ x \ln x - x \right]_{1}^{3}$ gives $(3 \ln 3 - 3) - (-1)$, which simplifies to $3 \ln 3 - 2$. Adding these two results produces $-\frac{4}{3} + \frac{8}{3} \ln 3$.

Factoring out $\frac{4}{3}$ allows the expression to be written as $\frac{4}{3} (2 \ln 3 - 1)$, which is equivalent to $\frac{4}{3} \ln \left( \frac{9}{e} \right)$. By equating this with the given form $\frac{m}{n} \ln \left( \frac{n^2}{e} \right)$, we identify $m = 4$ and $n = 3$. Given these coprime natural numbers, calculating $m^2 + n^2 - 5$ yields $16 + 9 - 5$, resulting in 20.

The approach to solving rational integrals where the numerator has a higher degree than the denominator relies on algebraic manipulation to simplify the fraction into easily integrable parts. By rewriting the term $t^{50}$ in the numerator as $t^{50} - 1 + 1$, the integral $\int_{0}^{\alpha} \frac{t^{50}}{1-t} \, dt$ can be expressed as $\int_{0}^{\alpha} \frac{t^{50}-1}{1-t} \, dt + \int_{0}^{\alpha} \frac{1}{1-t} \, dt$. The first term simplifies to $-\int_{0}^{\alpha} (1 + t + t^2 + \dots + t^{49}) \, dt$ because the quotient of $t^{50}-1$ and $1-t$ is $-(1 + t + t^2 + \dots + t^{49})$.

Evaluating these components individually reveals the final result. The integral of the series $1 + t + t^2 + \dots + t^{49}$ from $0$ to $\alpha$ is defined as $P_{50}(\alpha)$, and with the preceding negative sign, it becomes $-P_{50}(\alpha)$. The second integral, $\int_{0}^{\alpha} \frac{1}{1-t} \, dt$, evaluates to $[-\ln(1-t)]_{0}^{\alpha}$, which simplifies to $-\ln(1-\alpha)$. Given that $\beta = \log_{e}(1-\alpha)$, this term is equal to $-\beta$. Combining these values yields $-P_{50}(\alpha) - \beta$, which is equivalent to $-(\beta + P_{50}(\alpha))$.