Evaluate: $\int x^4 \log x \, dx$ OR Evaluate: $\int \frac{2x}{\sqrt[3]{x^2+1}} \, dx$

Evaluate: ∫ x⁴ log x dx OR Evaluate: ∫ frac2x√[3]x²+1 dx

Question

\int x^4 \log x \, dx

(OR)

\int \frac{2x}{\sqrt[3]{x^2+1}} \, dx

CBSECBSE Class XII Board 2020Subjective· 1mImportance★★★★★

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Part (a)Concept understanding — Integration by Parts

Integration by Parts

The idea: reverse the product rule

Some integrands are a product of two very different functions — $x e^x$ , $x\cos x$ , $\log x$ , $x\sin^{-1}x$ — where substitution gets you nowhere. Integration by parts is the tool for these. It comes straight from reversing the product rule for differentiation.

Starting from $\dfrac{d}{dx}(uv)=u\,v' + u'\,v$ and integrating both sides gives the working formula:

\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx.

In words: integral of (first $\times$ derivative-of-second) $=$ first $\times$ integral-of-second $-$ integral of (derivative-of-first $\times$ integral-of-second).

Choosing $u$ : the ILATE rule

The whole game is picking which factor is $u$ (to differentiate) and which is $dv$ (to integrate). Pick $u$ by ILATE — the first type that appears:

Inverse trig ( $\sin^{-1}x$ ), Logarithmic ( $\log x$ ), Algebraic ( $x^2$ ), Trigonometric ( $\sin x$ ), Exponential ( $e^x$ ).

Whatever comes first in ILATE becomes $u$ ; the rest is $dv$ . This makes the new integral $\int v\,du$ simpler than the one you started with.

Worked idea

For $\int x e^x\,dx$ : algebraic before exponential, so $u=x$ , $dv=e^x dx$ . Then $du=dx$ , $v=e^x$ :

$\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1)+C.$ …

Part (b)Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The chain rule differentiates composite functions: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given $\cos(x^2) \cdot 2x$ , find the original function. That's what u substitution does — it reverses the chain rule.

The Core Intuition

When an integral looks like "a function times the derivative of its inside," substitute the inside with $u$ and the derivative of the inside with $du$ . Consider:

\int 2x \cos(x^2) \, dx

Here $2x$ is the derivative of $x^2$ , and $x^2$ is the inside of $\cos(x^2)$ . Let $u = x^2$ , so $du = 2x \, dx$ :

\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C

Check: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ .

The Precise Statement

\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x), \, du = g'(x) \, dx

Valid provided $g$ is differentiable and the resulting integral in $u$ is simpler.

The Step-by-Step Method

Identify a function $g(x)$ whose derivative $g'(x)$ also appears (possibly up to a constant factor).
Set $u = g(x)$ , compute $du = g'(x) \, dx$ .
Rewrite the entire integral in $u$ and $du$ — every $x$ and $dx$ must be replaced.
Integrate with respect to $u$ .
Substitute back $u = g(x)$ .

Watch out

You cannot mix variables. If any $x$ remains after substitution, you chose the wrong $u$ (or must solve for $x$ in terms of $u$ — rare).

A Second Example (with a constant factor)

Evaluate $\int x \sqrt{x^2 + 1} \, dx$ . Let $u = x^2 + 1$ , so $x \, dx = \frac{1}{2} du$ :

\int \sqrt{u} \cdot \tfrac{1}{2} du = \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} + C = \tfrac{1}{3} (x^2 + 1)^{3/2} + C

When Does It Work?

When the integrand is something times the derivative of something inside. Common patterns:

$x \cdot f(x^2)$ — derivative of $x^2$ is $2x$ , so $u = x^2$
$e^{g(x)} \cdot g'(x)$ — derivative of $g(x)$ appears
$\frac{g'(x)}{g(x)}$ — leads to $\log|g(x)|$ …

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$\displaystyle\int x^4\log x\,dx=\tfrac{x^5}{5}\log x-\tfrac{x^5}{25}+C$ .
$\displaystyle\int\frac{2x}{\sqrt[3]{x^2+1}}\,dx=\tfrac32(x^2+1)^{2/3}+C$ .

Part (a)

Use integration by parts, $\displaystyle\int u\,dv=uv-\int v\,du$ , choosing $u=\log x$ (differentiates simply) and $dv=x^4\,dx$ , so $du=\tfrac1x\,dx$ and $v=\dfrac{x^5}{5}$ :

$\int x^4\log x\,dx=\frac{x^5}{5}\log x-\int\frac{x^5}{5}\cdot\frac{1}{x}\,dx=\frac{x^5}{5}\log x-\frac15\int x^4\,dx.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 84 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$\int \frac{3ax}{b^2 + c^2x^2} dx = A \log |b^2 + c^2x^2| + K$
›Reveal solutionSolution
The integral fits the pattern $\int \frac{du}{u} = \log|u| + C$ after a substitution. The constant $A$ turns out to be $\frac{3a}{2c^2}$ , which corresponds to option (D).

The problem gives you the result of an integral and asks you to identify the constant $A$ that makes the equation true. This is a classic "match the form" question — you don't need to guess; you just need to perform the integration carefully and compare.

The key insight is that the integrand $\frac{3ax}{b^2 + c^2x^2}$ is a rational function where the numerator is almost the derivative of the denominator. The derivative of $b^2 + c^2x^2$ is $2c^2x$ . Our numerator is $3ax$ , which is a constant multiple of $x$ . So a simple substitution $u = b^2 + c^2x^2$ will turn the integral into $\int \frac{du}{u}$ .

Let's work through it step by step.

Set up the substitution.

Let $u = b^2 + c^2x^2$ . Then $du = 2c^2x \, dx$ , so $x \, dx = \frac{du}{2c^2}$ .

Rewrite the integral in terms of $u$ .

The integral is $\int \frac{3ax}{b^2 + c^2x^2} \, dx = \int \frac{3a}{u} \cdot (x \, dx)$ .

Substitute $x \, dx = \frac{du}{2c^2}$ :

$\int \frac{3a}{u} \cdot \frac{du}{2c^2} = \frac{3a}{2c^2} \int \frac{du}{u}.$

Integrate. $\int \frac{du}{u} = \log|u| + C$ , so

$\frac{3a}{2c^2} \log|u| + C = \frac{3a}{2c^2} \log|b^2 + c^2x^2| + K,$
where $K$ is the constant of integration (we renamed $C$ to $K$ to match the problem).

Compare with the given form. …
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CBSE 2026Set CX1 mark
$\int x^{2}\tan(x^{3}+2)\,dx$
›Reveal solutionSolution

Substitute $u=x^{3}+2$ ; the integral becomes $\tfrac13\int\tan u\,du=\tfrac13\ln|\sec(x^{3}+2)|+C$ .

Concept: The factor $x^{2}$ is (up to a constant) the derivative of the inner function $x^{3}+2$ , so substitution works.

Let $u=x^{3}+2\Rightarrow du=3x^{2}\,dx\Rightarrow x^{2}\,dx=\dfrac{du}{3}$ .

…

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CBSE 2026Set A1 markMCQ
$\int\log x\,dx =$
(a) $\frac{1}{x} + k$
(b) $x\log x + k$
(c) $x\log x - x + k$
(d) $x\log x + x + k$
›Reveal solutionSolution

Integrate by parts: $\int\log x\,dx=x\log x-x+k$ .

Take $u=\log x$ and $dv=dx$ , so $du=\tfrac{1}{x}\,dx$ and $v=x$ :

…

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CBSE 2026Set A1 markMCQ
$\int\cos\sqrt{x}\,dx =$
(a) $\sin\sqrt{x} + \cos\sqrt{x} + k$
(b) $\frac{1}{2}(\sqrt{x}\sin\sqrt{x} - \cos\sqrt{x}) + k$
(c) $2(\sqrt{x}\sin\sqrt{x} + \cos\sqrt{x}) + k$
(d) $\sin\sqrt{x} + k$
›Reveal solutionSolution

Put $t=\sqrt{x}$ ; the integral becomes $2\int t\cos t\,dt=2(t\sin t+\cos t)+k$ .

Let $t=\sqrt{x}$ , so $x=t^2$ and $dx=2t\,dt$ . Then

$\int\cos\sqrt{x}\,dx=\int\cos t\,(2t\,dt)=2\int t\cos t\,dt$ .

Integrate $\int t\cos t\,dt$ by parts ( $u=t,\ dv=\cos t\,dt$ ): $=t\sin t-\int\sin t\,dt=t\sin t+\cos t$ .

…

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CBSE 2026Set A1 markMCQ
$\int e^x\left(\tan^{-1}x + \frac{1}{1+x^2}\right)dx =$
(a) $e^x\tan^{-1}x + k$
(b) $e^x \cdot \frac{1}{1+x^2} + k$
(c) $e^x + k$
(d) $\tan^{-1}x + k$
›Reveal solutionSolution

Recognise $\int e^x[f(x)+f'(x)]\,dx=e^x f(x)+k$ ; here $f(x)=\tan^{-1}x$ .

Note $\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}$ . So the integrand is $e^x\big[\tan^{-1}x+(\tan^{-1}x)'\big]$ , which matches the standard pattern

…

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CBSE 2026Set A1 markMCQ
$\int\frac{\tan(\sin^{-1}x)}{\sqrt{1-x^2}}\,dx =$
(a) $\log|\sec(\sin^{-1}x)| + k$
(b) $\log|\cos(\sin^{-1}x)| + k$
(c) $\tan(\sin^{-1}x) + k$
(d) $\log|\sin^{-1}x| + k$
›Reveal solutionSolution

With $u=\sin^{-1}x$ (so $du=\tfrac{dx}{\sqrt{1-x^2}}$ ) the integral is $\int\tan u\,du=\log|\sec u|+k$ .

Let $u=\sin^{-1}x$ . Then $du=\dfrac{dx}{\sqrt{1-x^2}}$ , so

…

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CBSE 2026Set A1 markMCQ
$\int\frac{dx}{e^x + e^{-x}} =$
(a) $\cot^{-1}(e^x) + k$
(b) $\tan^{-1}(e^x) + k$
(c) $\log|e^x + 1| + k$
(d) $\sin^{-1}(e^x) + k$
›Reveal solutionSolution

Substitute $t=e^x$ : $\int\tfrac{dx}{e^x+e^{-x}}=\int\tfrac{dt}{1+t^2}=\tan^{-1}(e^x)+k$ .

Multiply numerator and denominator by $e^x$ :

$\dfrac{1}{e^x+e^{-x}}=\dfrac{e^x}{e^{2x}+1}$ .

Let $t=e^x$ , $dt=e^x\,dx$ . Then

…

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CBSE 2026Set A1 markMCQ
$\int_0^1 x e^x\,dx =$
(a) $1$
(b) $0$
(c) $2$
(d) $-1$
›Reveal solutionSolution

By parts: $\int_0^1 xe^x\,dx=[(x-1)e^x]_0^1=1$ .

Take $u=x,\ dv=e^x\,dx$ , so $du=dx,\ v=e^x$ :

$\int x e^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x=(x-1)e^x$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\int e^x(\sin x-\cos x)\,dx$
(a) $-e^x\cos x+c$
(b) $e^x\sin x+c$
(c) $-e^x\sec x+c$
(d) $e^x\operatorname{cosec} x+c$
›Reveal solutionSolution

Recognising the standard form $\int e^x[f(x)+f'(x)]\,dx=e^xf(x)+c$ gives $-e^x\cos x+c$ .

There is a standard integration result:

$\int e^x\big[f(x)+f'(x)\big]\,dx=e^x f(x)+c$

Compare the integrand $\sin x-\cos x$ with $f(x)+f'(x)$ . Try $f(x)=-\cos x$ ; then $f'(x)=\sin x$ , so

$f(x)+f'(x)=-\cos x+\sin x=\sin x-\cos x$ …

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CBSE 2026Set ANNUAL1 markMCQ
$\int \sin(2x+3)\,dx =$
(a) $\cos(2x+3)+C$
(b) $-\dfrac{\cos(2x+3)}{2}+C$
(c) $\tan 2x + C$
(d) None of these
›Reveal solutionSolution

$\int \sin(ax+b)\,dx = -\dfrac{\cos(ax+b)}{a}+C$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\displaystyle\int_1^{e} \dfrac{(\log x)^2}{x}\, dx =$
(a) $\dfrac{1}{3}e^3$
(b) $\dfrac{1}{3}(e^3-1)$
(c) $\dfrac{1}{3}$
(d) None of these
›Reveal solutionSolution

Substitute $u = \log x$ , $du = dx/x$ , converting the limits from $x=1,e$ to $u=0,1$ .

Let $u=\log x \Rightarrow du = \dfrac{dx}{x}$ . When $x=1, u=0$ ; when $x=e, u=1$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\int e^x (\log \sec x + \tan x)\, dx =$
(a) $e^x + C$
(b) $e^x \tan x + C$
(c) $e^x(\log \sec x) + C$
(d) None of these
›Reveal solutionSolution

This is of the standard form $\int e^x[f(x)+f'(x)]\,dx = e^x f(x)+C$ .

Let $f(x) = \log\sec x$ . Then $f'(x) = \dfrac{\sec x\tan x}{\sec x} = \tan x$ .

…

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Q.Evaluate: ∫x4log⁡x dx\int x^4 \log x \, dx∫x4logxdx

Part (a)

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Q.Evaluate: $\int x^4 \log x \, dx$

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