Integrate the function $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$

Integrate the function √frac1-√(x)1+√(x)

Miscellaneous Exercise · Q19

\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}

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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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Rationalise the surd to $\dfrac{1-\sqrt x}{\sqrt{1-x}}$ , split it, and integrate: the result is $\sqrt{x-x^2}-2\sqrt{1-x}-\sin^{-1}\sqrt x+C$ .

1. Kill the nested root

Multiply numerator and denominator inside the square root by $1-\sqrt x$ (valid for $0<x<1$ , where $1-\sqrt x\ge0$ ):

\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}=\sqrt{\frac{(1-\sqrt x)^2}{(1+\sqrt x)(1-\sqrt x)}}=\frac{1-\sqrt x}{\sqrt{1-x}}.

Split it into two manageable pieces:

\frac{1-\sqrt x}{\sqrt{1-x}}=\frac{1}{\sqrt{1-x}}-\frac{\sqrt x}{\sqrt{1-x}}.

2. First integral

\int\frac{dx}{\sqrt{1-x}}=-2\sqrt{1-x}.

3. Second integral

Put $x=\sin^2\theta$ , so $dx=2\sin\theta\cos\theta\,d\theta$ , $\sqrt x=\sin\theta$ , $\sqrt{1-x}=\cos\theta$ :

$\int\frac{\sqrt x}{\sqrt{1-x}}\,dx=\int\frac{\sin\theta}{\cos\theta}\,2\sin\theta\cos\theta\,d\theta=2\int\sin^2\theta\,d\theta=\int(1-\cos2\theta)\,d\theta=\theta-\sin\theta\cos\theta.$ …

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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 32 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$\displaystyle\int \dfrac{\sqrt{\sqrt{x} + 1}}{\sqrt{x}}\, dx =$
›Reveal solutionSolution

Substitute $u=\sqrt{x}+1$ .

With $u=\sqrt{x}+1$ , $du=\dfrac{1}{2\sqrt{x}}\,dx$ , i.e. $\dfrac{dx}{\sqrt{x}}=2\,du$ . Then …
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KEAM 2026Set eng-2026-04184 marksMCQ
$\displaystyle\int \dfrac{\sec x\sqrt{\sec x}}{\sqrt{\sin x} + \sqrt{\cos x}}\, dx =$
›Reveal solutionSolution

Factor $\sqrt{\sin x}+\sqrt{\cos x}=\sqrt{\cos x}(\sqrt{\tan x}+1)$ to reduce the integrand to $\dfrac{\sec^2 x}{\sqrt{\tan x}+1}$ ; substituting $u=\sqrt{\tan x}$ gives option (A).

Write the denominator as $\sqrt{\sin x}+\sqrt{\cos x}=\sqrt{\cos x}\bigl(\sqrt{\tan x}+1\bigr)$ . The numerator is $\sec x\sqrt{\sec x}=\cos^{-3/2}x$ , so
$\dfrac{\cos^{-3/2}x}{\sqrt{\cos x}\,(\sqrt{\tan x}+1)}=\dfrac{\cos^{-2}x}{\sqrt{\tan x}+1}=\dfrac{\sec^2 x}{\sqrt{\tan x}+1}.$
Let $t=\tan x$ , $dt=\sec^2 x\,dx$ , then $u=\sqrt t$ , $t=u^2$ , $dt=2u\,du$ : …
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KEAM 2022Set eng-2022-P2-B14 marksMCQ
$\int \frac{1}{x^3}\sqrt{1 - \frac{1}{x^2}} \, dx =$
›Reveal solutionSolution
The integral equals $\frac13\left(1-\frac1{x^2}\right)^{3/2}+C$ .

Concept and Intuition

The derivative of $1-\frac1{x^2}$ is $\frac2{x^3}$ , which matches the $\frac1{x^3}$ factor outside the root, so a substitution linearizes the integral.

Step-by-Step Solution

Let $u=1-\frac1{x^2}$ .

$du=\frac{2}{x^3}\,dx\Rightarrow \frac{dx}{x^3}=\frac{du}{2}$ .

$\int \frac{1}{x^3}\sqrt{1-\frac1{x^2}}\,dx=\int \sqrt u\,\frac{du}{2}=\frac12\cdot\frac{u^{3/2}}{3/2}$ .

$=\frac12\cdot\frac23 u^{3/2}=\frac13 u^{3/2}=\frac13\left(1-\frac1{x^2}\right)^{3/2}+C$ . …
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KEAM 2024Set eng-2024-06094 marksMCQ
$\int \dfrac{dx}{x^2 (x^4 + 1)^{3/4}} =$
›Reveal solutionSolution

Factor $x^4$ out of the radical; with $u=1+x^{-4}$ , $du=-4x^{-5}dx$ , the integral becomes $-u^{1/4}=-\left(\frac{x^4+1}{x^4}\right)^{1/4}+C$ .

Write $(x^4+1)^{3/4}=x^3(1+x^{-4})^{3/4}$ , so

$\int\frac{dx}{x^2(x^4+1)^{3/4}}=\int\frac{dx}{x^5(1+x^{-4})^{3/4}}.$ …
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KEAM 2025Set eng-2025-04234 marksMCQ
$\int \dfrac{\sin^{-1} x}{\sqrt{1 - x^2}}\, dx =$
›Reveal solutionSolution
Substituting u=sin^{-1}x turns the integral into u du = (1/2)(sin^{-1}x)^2 + C.

Concept and Intuition

The factor 1/sqrt(1-x^2) is exactly the derivative of sin^{-1}x, so the substitution u = sin^{-1}x collapses the integral.

Step-by-Step Solution

Let u = sin^{-1}x, then du = dx/sqrt(1-x^2).

The integral becomes integral of u du. …
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KEAM 2025Set eng-2025-04294 marksMCQ
$\int \dfrac{dx}{\cos x \sqrt{2 \sin 2x}} =$
›Reveal solutionSolution

Writing $2\sin2x=4\sin x\cos x$ reduces the integrand to $\frac12\sec^2x\,(\tan x)^{-1/2}$ ; with $u=\tan x$ this integrates to $\sqrt{\tan x}+C$ .

Use $\sin 2x=2\sin x\cos x$ , so $2\sin 2x=4\sin x\cos x$ and $\sqrt{2\sin 2x}=2\sqrt{\sin x\cos x}$ . Then …
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KEAM 2022Set eng-2022-P2-B14 marksMCQ
$\int \frac{e^{\frac{1}{\sqrt{t}}}}{t\sqrt{t}} \, dt =$
›Reveal solutionSolution
The integral equals $-2e^{1/\sqrt t}+C$ .

Concept and Intuition

Recognize that the exponent's derivative appears (up to a constant) in the integrand, so a direct substitution works.

Step-by-Step Solution

Let $u=\frac1{\sqrt t}=t^{-1/2}$ .

$du=-\frac12 t^{-3/2}\,dt=-\frac12\cdot\frac{1}{t\sqrt t}\,dt$ , so $\frac{dt}{t\sqrt t}=-2\,du$ .

$\int \frac{e^{1/\sqrt t}}{t\sqrt t}\,dt=\int e^{u}(-2)\,du=-2e^{u}+C$ . …
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KEAM 2025Set eng-2025-04264 marksMCQ
$\displaystyle\int_0^{\pi/2} \dfrac{1}{1+\sin x}\, dx =$
›Reveal solutionSolution

Rationalise the denominator, integrate $\sec^2x-\sec x\tan x$ .

Multiply numerator and denominator by $1-\sin x$ :

$\frac{1}{1+\sin x}=\frac{1-\sin x}{1-\sin^2x}=\frac{1-\sin x}{\cos^2x}=\sec^2x-\sec x\tan x.$ …
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KEAM 2026Set eng-2026-04194 marksMCQ
$\int \frac{\sin(\cot^{-1}x)}{1 + x^2} dx$
›Reveal solutionSolution

Substitute $u=\cot^{-1}x$ so $du=-\tfrac{dx}{1+x^2}$ .

Let $u=\cot^{-1}x\Rightarrow du=-\dfrac{dx}{1+x^2}$ , i.e. $\dfrac{dx}{1+x^2}=-du$ . …
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KEAM 2026Set eng-2026-04184 marksMCQ
$\displaystyle\int \left(27x^3\left(1 - x^3\right)\right)^{\frac{2}{3}}\, dx =$
›Reveal solutionSolution

Simplify to $9x^2(1-x^3)^{2/3}$ , then substitute $u=1-x^3$ to obtain $-\dfrac{9}{5}(1-x^3)^{5/3}+C$ .

Since $\bigl(27x^3(1-x^3)\bigr)^{2/3}=27^{2/3}(x^3)^{2/3}(1-x^3)^{2/3}=9x^2(1-x^3)^{2/3}$ , let $u=1-x^3$ , $du=-3x^2\,dx$ , so $x^2\,dx=-\tfrac{du}{3}$ : …
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KEAM 2024Set eng-2024-06064 marksMCQ
$\int\frac{4x\cos\sqrt{4x^{2}+7}}{\sqrt{4x^{2}+7}}\,dx =$
›Reveal solutionSolution

With $u=\sqrt{4x^2+7}$ the integrand is exactly $\cos u\,du$ , giving $\sin\sqrt{4x^2+7}+C$ .

Let $u=\sqrt{4x^2+7}$ . Then $\frac{du}{dx}=\frac{8x}{2\sqrt{4x^2+7}}=\frac{4x}{\sqrt{4x^2+7}}$ , so $du=\frac{4x}{\sqrt{4x^2+7}}\,dx$ . …
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KEAM 2024Set eng-2024-06054 marksMCQ
$\int \frac{dx}{x^8 \left(\frac{1}{x^7} + 1\right)^{2/3}}$
›Reveal solutionSolution

The substitution $u=x^{-7}+1$ reduces it to $-\frac17\int u^{-2/3}\,du$ .

Let $u=\dfrac{1}{x^7}+1$ . Then $du=-\dfrac{7}{x^8}\,dx$ , so $\dfrac{dx}{x^8}=-\dfrac{du}{7}$ . The integral becomes …
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Q.Integrate the function $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$

Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The Core Intuition

The Precise Statement

The Step-by-Step Method

A Second Example (with a constant factor)

When Does It Work?

1. Kill the nested root

2. First integral

3. Second integral

Method: Rationalise a nested surd, then split and use a trig substitution

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.Integrate the function 1−x1+x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}1+x​1−x​​​

1. Kill the nested root

2. First integral

3. Second integral

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Q.Integrate the function $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$

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