Indefinite Integrals – JEE Main MathematicsPractice Questions & PYQs

The integral splits naturally into two distinct components, $I_1$ and $I_2$, based on the structure of the integrand. For $I_1 = \int \frac{x^{2}-1}{(x^{4}+3x^{2}+1)\tan^{-1}\left(\frac{x^{2}+1}{x}\right)} dx$, the substitution $t = \tan^{-1}\left(\frac{x^{2}+1}{x}\right)$ simplifies the expression completely, as $dt = \frac{x^{2}-1}{x^{4}+3x^{2}+1} dx$. This results in the term $\int \frac{1}{t} dt = \log\left|\tan^{-1}\left(\frac{x^{2}+1}{x}\right)\right|$, which identifies the coefficient $\alpha = 1$.

The second component, $I_2 = \int \frac{1}{x^{4}+3x^{2}+1} dx$, is evaluated by expressing the numerator as $\frac{1}{2}((x^{2+}1) - (x^{2-}1))$. Dividing the numerator and denominator by $x^2$ transforms this into $\frac{1}{2} \int \frac{1+1/x^{2}}{(x-1/x)^{2}+5} dx - \frac{1}{2} \int \frac{1-1/x^{2}}{(x+1/x)^{2}+1} dx$. These integrals correspond to standard forms, yielding $\frac{1}{2\sqrt{5}} \tan^{-1}\left(\frac{x-1/x}{\sqrt{5}}\right) - \frac{1}{2} \tan^{-1}\left(x+1/x\right)$. Recognizing $x - 1/x = \frac{x^{2-}1}{x}$ and $x + 1/x = \frac{x^{2+}1}{x}$, this matches the required form $\beta \tan^{-1}\left(\frac{\gamma(x^{2}-1)}{x}\right) + \delta \tan^{-1}\left(\frac{x^{2}+1}{x}\right)$, where $\beta = \frac{1}{2\sqrt{5}}$, $\gamma = \frac{1}{\sqrt{5}}$, and $\delta = -\frac{1}{2}$.

Substituting these values into the final expression $10(\alpha + \beta \gamma + \delta)$ yields $10(1 + (\frac{1}{2\sqrt{5}} \cdot \frac{1}{\sqrt{5}}) - \frac{1}{2})$. This calculation simplifies to $10(1 + \frac{1}{10} - \frac{1}{2})$, which equals $10(\frac{10+1-5}{10})$, resulting in a final value of $6$.

Multiplying the numerator and denominator of the expression $\int \frac{2 e^{x}+3 e^{-x}}{4 e^{x}+7 e^{-x}} dx$ by $e^x$ results in the integral $\int \frac{2 e^{2x}+3}{4 e^{2x}+7} dx$. To solve this, we can decompose the numerator by expressing it as a linear combination of the denominator and its derivative. Given that the derivative of $4 e^{2x}+7$ is $8 e^{2x}$, we set up the equation $2 e^{2x}+3 = A(8 e^{2x}) + B(4 e^{2x}+7)$. By comparing coefficients, we obtain $8A + 4B = 2$ and $7B = 3$, which yield $A = \frac{1}{28}$ and $B = \frac{3}{7}$.

Substituting these values transforms the integral into the sum $\int \frac{A(8 e^{2x})}{4 e^{2x}+7} dx + \int B dx$, evaluating to $A \ln|4 e^{2x}+7| + Bx + C$. Substituting the values for $A$ and $B$, we have $\frac{1}{28} \ln|4 e^{2x}+7| + \frac{3}{7} x + C$. Expanding the logarithm as $\ln|e^x(4 e^x+7 e^{-x})| = x + \ln|4 e^x+7 e^{-x}|$, the expression becomes $\frac{1}{28} x + \frac{1}{28} \ln|4 e^x+7 e^{-x}| + \frac{3}{7} x + C$. Combining the linear terms $\frac{1}{28} x + \frac{12}{28} x$ results in $\frac{13}{28} x$, allowing the entire expression to be written as $\frac{1}{14}(\frac{13}{2} x + \frac{1}{2} \ln|4 e^x+7 e^{-x}|) + C$. Comparing this to the target form reveals $u = \frac{13}{2}$ and $v = \frac{1}{2}$, and summing these constants gives $u+v = 7$.

Dividing the numerator and denominator of the integral by $\cos^3 x$ transforms the expression into $\int \frac{\tan x \sec^2 x}{\tan^3 x + 1} \, dx$. Applying the substitution $t = \tan x$, which implies $dt = \sec^2 x \, dx$, simplifies the integral to $\int \frac{t}{t^3 + 1} \, dt$. Factoring the denominator as $(t+1)(t^2 - t + 1)$ allows for a partial fraction decomposition of the form $\frac{A}{t+1} + \frac{Bt + C}{t^2 - t + 1}$.

Solving for the coefficients yields $A = -1/3$, $B = 1/3$, and $C = 1/3$. The integral then splits into logarithmic and inverse tangent components, resulting in the antiderivative:
$-\frac{1}{3} \ln|t+1| + \frac{1}{6} \ln|t^2 - t + 1| + \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2t-1}{\sqrt{3}}\right) + C$

Substituting $t = \tan x$ back into this expression identifies the constants as $\alpha = -1/3$, $\beta = 1/6$, and $\gamma = 1/\sqrt{3}$. Evaluating the final expression $18(\alpha + \beta + \gamma^2)$ requires computing $18(-1/3 + 1/6 + 1/3)$, which results in 3.

The integral becomes manageable by manipulating the denominator to reveal a term whose derivative appears in the numerator. Factoring $x^7$ out of the expression $x^2 + 1 + 2x^7$ yields $x^7(x^{-5} + x^{-7} + 2)$, which effectively transforms the integral into $\int \frac{5x^{-6} + 7x^{-8}}{(x^{-5} + x^{-7} + 2)^2} \, dx$ after accounting for the division by $x^{14}$. Defining $z = x^{-5} + x^{-7} + 2$ results in $dz = -(5x^{-6} + 7x^{-8}) \, dx$, simplifying the integrand to $-\int z^{-2} \, dz$.

Integrating this expression yields $\frac{1}{z} + C$, which is equivalent to $f(x) = \frac{1}{x^{-5} + x^{-7} + 2} + C$, or $f(x) = \frac{x^7}{x^2 + 1 + 2x^7} + C$. Applying the condition $f(0) = 0$ confirms that $C = 0$. Consequently, $f(1) = \frac{1^7}{1^2 + 1 + 2(1^7)} = \frac{1}{1 + 1 + 2} = \frac{1}{4}$, which implies that $K = 4$.

This integral is solved by recognizing that the expression inside the cosine function is functionally linked to the denominator, which allows for a substitution that simplifies the entire expression. By observing that the denominator $4x^{2-}4x+6$ can be rewritten as $(2x-1)^{2+}5$, we can define a new variable $u = \sqrt{(2x-1)^{2+}5}$ to handle the complexity of the integrand.

Differentiating $u^2 = (2x-1)^{2+}5$ with respect to $x$ yields $2u \, du = 2(2x-1) \cdot 2 \, dx$, which simplifies to $u \, du = 2(2x-1) \, dx$, or $(2x-1) \, dx = \frac{1}{2}u \, du$. Substituting these components into the original integral transforms it into $\int \frac{\cos(u)}{u} \cdot \frac{1}{2}u \, du$. This cancellation of $u$ leaves us with the much simpler integral $\frac{1}{2} \int \cos(u) \, du$, which integrates to $\frac{1}{2} \sin(u) + c$. Reversing the substitution leads directly to the final result of $\frac{1}{2}\sin\sqrt{(2x-1)^{2+}5}+c$.

By applying the logarithmic property $a \log_e b = \log_e b^a$ and the identity $e^{\log_e u} = u$, the exponential terms in the integrand simplify directly into algebraic form. The numerator terms transform as $e^{3 \log_e (2x)} = (2x)^3 = 8x^3$ and $5 e^{2 \log_e (2x)} = 5(2x)^2 = 20x^2$, while the denominator terms simplify to $e^{4 \log_e x} = x^4$, $5 e^{3 \log_e x} = 5x^3$, and $-7 e^{2 \log_e x} = -7x^2$. This yields the simplified integrand $\frac{8x^3 + 20x^2}{x^4 + 5x^3 - 7x^2}$, which can be reduced by factoring out $4x^2$ from the numerator and $x^2$ from the denominator to obtain $\int \frac{4(2x+5)}{x^{2+}5x-7} dx$. Recognizing that the numerator $4(2x+5)dx$ is proportional to the derivative of the denominator $x^{2+}5x-7$, the substitution $t = x^{2+}5x-7$ with $dt = (2x+5)dx$ leads to $\int \frac{4}{t} dt$, resulting in $4 \log_e |x^{2+}5x-7| + c$.

Integration by parts is the most effective approach here because we can manipulate the integrand into a form involving a function and its derivative. By expressing the squared term as the product $\left(\frac{x}{\cos x}\right) \cdot \frac{x\cos x}{(x\sin x + \cos x)^2}$, we define $u = \frac{x}{\cos x}$ and $dv = \frac{x\cos x}{(x\sin x + \cos x)^2} \, dx$. Differentiating $u$ yields $du = \frac{\cos x + x\sin x}{\cos^2 x} \, dx$, and integrating $dv$ yields $v = -\frac{1}{x\sin x + \cos x}$, noting that the numerator $x\cos x$ is the derivative of the denominator $x\sin x + \cos x$. Substituting these into the formula $\int u \, dv = uv - \int v \, du$ produces $-\frac{x}{\cos x(x\sin x + \cos x)} - \int -\left(\frac{1}{x\sin x + \cos x}\right) \left(\frac{\cos x + x\sin x}{\cos^2 x}\right) \, dx$. After canceling the common term $x\sin x + \cos x$ in the integral, we are left with $\int \sec^2 x \, dx$, which evaluates to $\tan x$. This provides the final result of $\tan x - \frac{x\sec x}{x\sin x + \cos x} + C$.

The sum of the exponents in the denominator, $-8/7 - 6/7$, equals $-2$, which identifies that the integral can be resolved by substituting the ratio of the linear factors. Defining $t = \frac{x+4}{x-3}$ leads to the derivative $dt = -\frac{7}{(x-3)^2} dx$, which allows the differential $dx$ to be expressed as $-\frac{1}{7} dt (x-3)^2$.

Rearranging the integral to accommodate this substitution provides:
$\int \frac{dx}{\left(\frac{x+4}{x-3}\right)^{8/7} (x-3)^2}$
Substituting $t$ and $dt$ into the expression transforms it into a basic power rule form:
$-\frac{1}{7} \int t^{-8/7} dt = -\frac{1}{7} \left( \frac{t^{-1/7}}{-1/7} \right) + C = t^{-1/7} + C$
Substituting the original definition of $t$ back into the result yields $\left(\frac{x+4}{x-3}\right)^{-1/7} + C$, which simplifies to $\left(\frac{x-3}{x+4}\right)^{1/7} + C$.

Expressing the trigonometric functions in terms of $\tan \theta$ by utilizing double angle identities reveals the underlying structure of this integral. Because $\frac{1}{\cos^2 \theta}$ equates to $\sec^2 \theta$, substituting $t = \tan \theta$ with $dt = \sec^2 \theta d\theta$ proves highly effective. Expanding the denominator terms using $\tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta}$ and $\sec 2\theta = \frac{1+\tan^2\theta}{1-\tan^2\theta}$ results in a simplified form of $\frac{(1+\tan\theta)^2}{1-\tan^2\theta}$, allowing the original integral to be rewritten as $\int \frac{1-t^2}{(1+t)^2} dt$.

Factoring the numerator $1-t^2$ as $(1-t)(1+t)$ permits canceling one $(1+t)$ term, simplifying the integrand to $\frac{1-t}{1+t}$. Expressing the numerator as $2-(1+t)$ transforms the integral into the form $\int \left( \frac{2}{1+t} - 1 \right) dt$. Evaluating this yields $2 \ln |1+t| - t + C$. Substituting $t = \tan \theta$ back into this result provides the final expression $2 \ln |1+\tan\theta| - \tan\theta + C$, from which it is evident that $\lambda = -1$ and $f(\theta) = 1 + \tan\theta$.

The expression $\sec x + \tan x$ can be simplified using trigonometric identities as follows:
$\sec x + \tan x = \frac{1 + \sin x}{\cos x} = \frac{\left(\cos \frac{x}{2} + \sin \frac{x}{2}\right)^2}{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}} = \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$
Substituting this result into the derivative function gives $f'(x) = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right) = \frac{\pi}{4} + \frac{x}{2}$ for the given domain. Integrating both sides with respect to $x$ yields $f(x) = \int \left(\frac{\pi}{4} + \frac{x}{2}\right) dx = \frac{\pi x}{4} + \frac{x^2}{4} + C$.

Using the condition $f(0) = 0$, the constant of integration $C$ evaluates to $0$, defining the function as $f(x) = \frac{\pi x + x^2}{4}$. Calculating the value of $f(1)$ results in $f(1) = \frac{\pi(1) + 1^2}{4} = \frac{\pi + 1}{4}$.

Solving this integral requires identifying a suitable substitution by recognizing that the differential of $\csc^6 x$ leads to terms involving $\sin^{-7} x \cos x$. By factoring $\sin^6 x$ out of the term $(1+\sin^6 x)^{2/3}$ in the denominator, the integrand can be rewritten as $\int \sin^{-7} x (1 + \sin^{-6} x)^{-2/3} \cos x \, dx$. Setting $u = 1 + \sin^{-6} x$ results in $du = -6 \sin^{-7} x \cos x \, dx$, which transforms the expression into $-\frac{1}{6} \int u^{-2/3} \, du$. Integrating this yields $-\frac{1}{6} \times 3u^{1/3} + c = -\frac{1}{2}(1 + \sin^{-6} x)^{1/3} + c$.

Algebraically simplifying the result leads to $-\frac{1}{2} \left( \frac{1+\sin^6 x}{\sin^6 x} \right)^{1/3} + c = -\frac{1}{2\sin^2 x}(1+\sin^6 x)^{1/3} + c$. Comparing this result with the structure $f(x)(1+\sin^6 x)^{1/\lambda} + c$ indicates that $\lambda = 3$ and $f(x) = -\frac{1}{2\sin^2 x}$. Evaluating this function at $x = \pi/3$ gives $f(\pi/3) = -\frac{1}{2(\sin(\pi/3))^2} = -\frac{1}{2(3/4)} = -\frac{2}{3}$. Multiplying this by $\lambda$ yields $3 \times (-2/3) = -2$.

To evaluate this integral, begin by strategically rearranging the algebraic expression $e^{2x}+2e^{x}-e^{-x}-1$ to identify a pattern suitable for integration by parts. By factoring and grouping, this expression can be rewritten as $(e^{x}+1)(e^{x}-e^{-x}) + e^{x}$, allowing the integral to be split into $\int (e^{x}+1)(e^{x}-e^{-x}) e^{e^{x}+e^{-x}} dx + \int e^{x} e^{e^{x}+e^{-x}} dx$. Focus on the first part of this split and apply the integration by parts method by choosing $u = e^{x}+1$ and $dv = (e^{x}-e^{-x}) e^{e^{x}+e^{-x}} dx$. Since the integral of $dv$ is $e^{e^{x}+e^{-x}}$, applying the formula $\int u dv = uv - \int v du$ transforms the first integral into $(e^{x}+1)e^{e^{x}+e^{-x}} - \int e^{x} e^{e^{x}+e^{-x}} dx$. When this result is added to the second term of the original integral, $\int e^{x} e^{e^{x}+e^{-x}} dx$, the two integrals cancel out completely, leaving the final expression $(e^{x}+1)e^{e^{x}+e^{-x}}+c$. Comparing this to the form $g(x)e^{e^{x}+e^{-x}}+c$ reveals that $g(x) = e^{x}+1$, which means $g(0) = e^{0}+1 = 2$.

The fundamental strategy for solving this integral relies on converting the trigonometric expression into a standard algebraic form through substitution. Since the denominator contains both $\cos^2 \theta$ and $\sin \theta$, using the identity $\cos^2 \theta = 1 - \sin^2 \theta$ allows us to rewrite the denominator $5 + 7 \sin \theta - 2 \cos^2 \theta$ as $2 \sin^2 \theta + 7 \sin \theta + 3$. By defining $t = \sin \theta$, we get $dt = \cos \theta d\theta$, which transforms the integral into the rational form $\int \frac{dt}{2t^2 + 7t + 3}$.

Factoring the quadratic denominator yields $(2t+1)(t+3)$, enabling the use of partial fractions to decompose the integrand into $\frac{1}{5} \left( \frac{2}{2t+1} - \frac{1}{t+3} \right)$. Integrating each term results in $\frac{1}{5} \ln |2t+1| - \frac{1}{5} \ln |t+3| + C$, which simplifies to $\frac{1}{5} \ln \left| \frac{2 \sin \theta+1}{\sin \theta+3} \right| + C$. Mapping this result to the given expression $A \ln |B(\theta)| + C$, we identify $A = \frac{1}{5}$ and $B(\theta) = \frac{2 \sin \theta+1}{\sin \theta+3}$. Consequently, the ratio $\frac{B(\theta)}{A}$ is obtained by dividing $\frac{2 \sin \theta+1}{\sin \theta+3}$ by $\frac{1}{5}$, which simplifies to $\frac{5(2 \sin \theta+1)}{\sin \theta+3}$.

Using the substitution $x = \tan^{2} \theta$ allows $dx = 2 \tan \theta \sec^{2} \theta d\theta$ and simplifies the inverse sine term to $\theta$, transforming the integral into
$\int \theta (2 \tan \theta \sec^{2} \theta) d\theta$
Applying integration by parts with $u = \theta$ and $dv = 2 \tan \theta \sec^{2} \theta d\theta$ leads to $\theta \tan^{2} \theta - \int \tan^{2} \theta d\theta$. Because $\tan^{2} \theta = \sec^{2} \theta - 1$, this simplifies further to $\theta(1 + \tan^{2} \theta) - \tan \theta + C$. Expressing this result in terms of $x$ gives $(1 + x) \tan^{-1}(\sqrt{x}) - \sqrt{x} + C$, which identifies $A(x) = x+1$ and $B(x) = -\sqrt{x}$.

The substitution method simplifies complex radical integrands by transforming them into rational expressions. To compute $f(3)-f(1)$, evaluate the definite integral $\int_1^3 \frac{\sqrt{x}}{(1+x)^2} \, dx$ by setting $t = \sqrt{x}$, which implies $x = t^2$ and $dx = 2t \, dt$, adjusting the limits of integration from 1 to $\sqrt{3}$. This change yields the integral $\int_1^{\sqrt{3}} \frac{2t^2}{(1+t^2)^2} \, dt$.

Applying integration by parts with $u = t$ and $dv = \frac{2t}{(1+t^2)^2} \, dt$ results in $du = dt$ and $v = -\frac{1}{1+t^2}$, leading to the antiderivative $-\frac{t}{1+t^2} + \tan^{-1}(t)$. Evaluating this expression between the limits 1 and $\sqrt{3}$ produces $\left(-\frac{\sqrt{3}}{4} + \frac{\pi}{3}\right) - \left(-\frac{1}{2} + \frac{\pi}{4}\right)$. Simplifying these values gives $\frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4}$.

Integration of complex rational expressions often relies on algebraic rearrangement to reveal a hidden derivative-function relationship. By strategically dividing terms by a specific power of $x$, a complicated integrand can be transformed into a straightforward power-rule substitution problem.

Dividing both the numerator $3x^{13}+2x^{11}$ and the base of the denominator $(2x^{4+}3x^{2+}1)^4$ by $x^{16}$ effectively scales the expression without changing its value. This transformation leads to the integral form $\int \frac{3x^{-3}+2x^{-5}}{(2+3x^{-2}+x^{-4})^4} dx$. By defining $t = 2+3x^{-2}+x^{-4}$, the differential $dt$ becomes $(-6x^{-3}-4x^{-5}) dx$, which is simply $-2(3x^{-3}+2x^{-5}) dx$. The integral then becomes:
$\int -\frac{1}{2t^4} dt = -\frac{1}{2} \left(\frac{t^{-3}}{-3}\right) + C = \frac{1}{6t^3} + C$
Substituting the original polynomial expression back in for $t$ yields $\frac{1}{6(2+3x^{-2}+x^{-4})^3} + C$. Simplifying this fraction by clearing the negative powers results in the final form:
$\frac{x^{12}}{6(2x^{4+}3x^{2+}1)^3} + C$

The method of substitution provides a clear pathway for solving integrals that feature algebraic radicals, particularly when the integrand contains a derivative relationship with the variable. By rewriting the expression as $\int \frac{\sqrt{1-x^2}}{x} \cdot \frac{1}{x^3} dx$, we can observe that $\frac{\sqrt{1-x^2}}{x}$ is equivalent to $\sqrt{\frac{1}{x^2}-1}$. Setting $t = \frac{1}{x^2}-1$ allows us to define the differential $dt = -\frac{2}{x^3} dx$, which elegantly transforms the integral into $-\frac{1}{2} \int \sqrt{t} \, dt$.