Correct Answer is : 2(x−tan⁡−1x)+c2≤ft({x}- ^{-1} {x})+c2(x−tan−1x)+c, where c is a constant of integration.

≤ft(2 {x}- ^{-1} {x})+c, where c is a constant of integration.

2≤ft({x}- ^{-1} {x})+c, where c is a constant of integration.

≤ft(2 {x}+ ^{-1} {x})+c, where c is a constant of integration.

2≤ft({x}+ ^{-1} {x})+c, where c is a constant of integration.

Correct Answer is : 2(x−tan⁡−1x)+c2≤ft({x}- ^{-1} {x})+c2(x−tan−1x)+c, where c is a constant of integration.

Test Index

MHTCET 10 May 2024 Shift 1 Paper

Question : 131 of 150

Marks: +1, -0

\int \frac{\sqrt{x}}{x+1} d x=

$\left(2 \sqrt{x}-\tan ^{-1} \sqrt{x}\right)+c$ , where c is a constant of integration.
$2\left(\sqrt{x}-\tan ^{-1} \sqrt{x}\right)+c$ , where c is a constant of integration.
$\left(2 \sqrt{x}+\tan ^{-1} \sqrt{x}\right)+c$ , where c is a constant of integration.
$2\left(\sqrt{x}+\tan ^{-1} \sqrt{x}\right)+c$ , where c is a constant of integration.

Solution:

Concept:

Substitute

\sqrt{x}=t

to convert the irrational integrand into a rational function in

t

Explanation:

Let

\sqrt{x}=t

, so

x=t^2

and

dx = 2t\,dt

Then the integral becomes

I = \int \frac{t}{t^2+1} \cdot 2t\,dt = \int \frac{2t^2}{t^2+1}\,dt.

Rewrite the numerator as

t^2+1-1

to split the fraction:

I = 2\int \left(1 - \frac{1}{t^2+1}\right)dt.

Now integrate term by term:

I = 2\left(t - \tan^{-1}t\right) + c.

Substitute back

t = \sqrt{x}

I = 2\left(\sqrt{x} - \tan^{-1}\sqrt{x}\right) + c.

This matches option B.

Answer:

2\left(\sqrt{x} - \tan^{-1}\sqrt{x}\right) + c

, where

c

is a constant of integration.

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