Find: $\int \frac{1}{\sqrt{x}(\sqrt{x} + 1)(\sqrt{x} + 2)} dx$

> [!TLDR] > The integral simplifies by substituting $t = \sqrt{x}$, turning it into a rational function that decomposes into partial fractions. The final result is $\boxed{2\log\left|\frac{\sqrt{x}+1}{\sqrt{x}+2}\right| + C}$. ## Why substitution works here The integrand is a messy combination of $\sqrt{x}$ and $x$ in the denominator. The key insight: whenever you see $\sqrt{x}$ repeated in multiple places, let $t = \sqrt{x}$. This turns every $\sqrt{x}$ into $t$, and $dx$ becomes $2t\,dt$, which often cancels nicely with the $\frac{1}{\sqrt{x}}$ factor already present. Notice the $\frac{1}{\sqrt{x}}$ at the front — that's a hint. After substitution, it will combine with $dx$ to give a clean $2\,dt$, leaving a purely rational function in $t$. ## Step-by-step solution 1. **Set the substitution** Let $t = \sqrt{x}$. Then $x = t^2$, and differentiating: $$dx = 2t\,dt$$ 2. **Rewrite the integral** Replace every $\sqrt{x}$ with $t$, and $dx$ with $2t\,dt$: $$\int \frac{1}{\sqrt{x}(\sqrt{x} + 1)(\sqrt{x} + 2)} dx = \int \frac{1}{t(t+1)(t+2)} \cdot 2t\,dt$$ The $t$ in the numerator cancels with the $t$ in the denominator: $$= \int \frac{2}{(t+1)(t+2)}\,dt$$ > [!TIP] > The cancellation is the whole point — the $\frac{1}{\sqrt{x}}$ factor was designed to make this work. If it weren't there, the substitution would still be possible but messier. 3. **Partial fraction decomposition** We need to break $\frac{2}{(t+1)(t+2)}$ into simpler pieces. Write: $$\frac{2}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2}$$ Multiply through by $(t+1)(t+2)$: $$2 = A(t+2) + B(t+1)$$ Solve for $A$ and $B$. A fast method: - Set $t = -1$: $2 = A(1) + B(0) \implies A = 2$ …

Find: ∫ frac1√(x)(√(x) + 1)(√(x) + 2) dx

Question

\int \frac{1}{\sqrt{x}(\sqrt{x} + 1)(\sqrt{x} + 2)} dx

CBSECBSE Class XII Board 2023Subjective· 3mImportance★★★★★

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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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The integral simplifies by substituting $t = \sqrt{x}$ , turning it into a rational function that decomposes into partial fractions. The final result is $\boxed{2\log\left|\frac{\sqrt{x}+1}{\sqrt{x}+2}\right| + C}$ .

Why substitution works here

The integrand is a messy combination of $\sqrt{x}$ and $x$ in the denominator. The key insight: whenever you see $\sqrt{x}$ repeated in multiple places, let $t = \sqrt{x}$ . This turns every $\sqrt{x}$ into $t$ , and $dx$ becomes $2t\,dt$ , which often cancels nicely with the $\frac{1}{\sqrt{x}}$ factor already present.

Notice the $\frac{1}{\sqrt{x}}$ at the front — that's a hint. After substitution, it will combine with $dx$ to give a clean $2\,dt$ , leaving a purely rational function in $t$ .

Step-by-step solution

Set the substitution Let $t = \sqrt{x}$ . Then $x = t^2$ , and differentiating:

dx = 2t\,dt

Rewrite the integral Replace every $\sqrt{x}$ with $t$ , and $dx$ with $2t\,dt$ :

\int \frac{1}{\sqrt{x}(\sqrt{x} + 1)(\sqrt{x} + 2)} dx = \int \frac{1}{t(t+1)(t+2)} \cdot 2t\,dt

The $t$ in the numerator cancels with the $t$ in the denominator:

= \int \frac{2}{(t+1)(t+2)}\,dt

Tip

The cancellation is the whole point — the $\frac{1}{\sqrt{x}}$ factor was designed to make this work. If it weren't there, the substitution would still be possible but messier.

Partial fraction decomposition We need to break $\frac{2}{(t+1)(t+2)}$ into simpler pieces. Write:

\frac{2}{(t+1)(t+2)} = \frac{A}{t+1} + \frac{B}{t+2}

Multiply through by $(t+1)(t+2)$ :

2 = A(t+2) + B(t+1)

Solve for $A$ and $B$ . A fast method:

Set $t = -1$ : $2 = A(1) + B(0) \implies A = 2$ …

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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 44 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
$\int\frac{\tan(\sin^{-1}x)}{\sqrt{1-x^2}}\,dx =$
(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 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
$\displaystyle\int \dfrac{1}{x(1+\log x)}\,dx$
(a) $x + \log x + c$
(b) $|x + \log x| + c$
(c) $\log|1 + \log x| + c$
(d) $\log(1+x) + c$
›Reveal solutionSolution

Substitute $u=1+\log x$ so $du=\dfrac{dx}{x}$ , turning the integral into $\int\dfrac{du}{u}$ .

$I=\displaystyle\int\dfrac{1}{x(1+\log x)}\,dx$

Let $u=1+\log x \Rightarrow du=\dfrac{1}{x}dx$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{\sin^6 x}{\cos^8 x}\,dx$
›Reveal solutionSolution

Rewrite the integrand as $\tan^6x\sec^2x$ and substitute $t=\tan x$ .

$I=\displaystyle\int\dfrac{\sin^6x}{\cos^8x}\,dx=\int\dfrac{\sin^6x}{\cos^6x}\cdot\dfrac{1}{\cos^2x}\,dx=\int\tan^6x\sec^2x\,dx$

Let $t=\tan x \Rightarrow dt=\sec^2x\,dx$ . …

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CBSE 2026Set ANNUAL1 markMCQ
$Q.\int x^2 e^{x^3} dx equals:$
(a)
(i) \frac{e^{x^3}}{3} + c
(b)
(ii) 3e^{x^3} + c
(c)
(iii) \frac{e^{x^2}}{3} + c
(d)
(iv) \frac{1}{2}e^{x^2} + c
›Reveal solutionSolution
$\displaystyle\int x^2 e^{x^3}\,dx=\dfrac{e^{x^3}}{3}+c$ — option (i).

Concept. Substitution ( $u$ -substitution): choose $u$ whose derivative already appears (up to a constant) in the integrand.

Steps.

Let $u=x^3$ , then $du=3x^2\,dx$ , i.e. $x^2\,dx=\dfrac{1}{3}\,du$ . …
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CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{\cos x}{1+\sin^2 x}\, dx =$
(a) $-\tan^{-1}(\sin x) + c$
(b) $\tan^{-1}(\cos x) + c$
(c) $\tan^{-1}(\sin x) + c$
(d) $-\tan^{-1}(\cos x) + c$
›Reveal solutionSolution

A direct substitution u = sin x reduces this to the standard ∫du/(1+u²) form.

Let $u=\sin x$ , so $du = \cos x\,dx$ .

…

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CBSE 2025Set ANNUAL1 mark
$Q. Evaluate \int 2x sin(1 - x^2) dx.$
›Reveal solutionSolution

Substitute $u = 1-x^2$ so $du = -2x\,dx$ , turning the integral into a simple $\int \sin u\,du$ .

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

…

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CBSE 2025Set ANNUAL1 mark
$\displaystyle\int_{2}^{\sqrt3} \dfrac{x\,dx}{x^2+1}$
›Reveal solutionSolution

Substitute $u=x^2+1$ so the integrand becomes $\frac12\int\frac{du}{u}$ , then apply the given limits (2 to $\sqrt3$ ) directly.

$\int_{2}^{\sqrt3}\dfrac{x\,dx}{x^2+1}$

Let $u = x^2+1$ , so $du = 2x\,dx$ , i.e. $x\,dx = \dfrac{du}{2}$ .

$\int\dfrac{x\,dx}{x^2+1} = \dfrac12\int\dfrac{du}{u} = \dfrac12\ln|u| = \dfrac12\ln(x^2+1)$

Applying the limits exactly as given (lower limit $x=2$ , upper limit $x=\sqrt3$ ):

$\left[\dfrac12\ln(x^2+1)\right]_{2}^{\sqrt3} = \dfrac12\ln(3+1) - \dfrac12\ln(4+1) = \dfrac12\ln4 - \dfrac12\ln5 = \dfrac12\ln\dfrac{4}{5}$

…

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Q.Find: ∫1x(x+1)(x+2)dx\int \frac{1}{\sqrt{x}(\sqrt{x} + 1)(\sqrt{x} + 2)} dx∫x​(x​+1)(x​+2)1​dx

Why substitution works here

Step-by-step solution

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Q.Find: $\int \frac{1}{\sqrt{x}(\sqrt{x} + 1)(\sqrt{x} + 2)} dx$

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