^{-1}≤ft({2 x}{1+x²}) d x=

Correct Answer is : 2xtan⁡−1x−log⁡(1+x2)+c2 x ^{-1} x- ≤ft(1+x²)+c2xtan−1x−log(1+x2)+c, where c is a constant of integration.

2 x ^{-1} x- ≤ft(1+x²)+c, where c is a constant of integration.

2≤ft(x ^{-1} x- ≤ft(1+x²))+c, where c is a constant of integration.

x ^{-1} x+ ≤ft(1+x²)+c, where c is a constant of integration.

2≤ft(x ^{-1} x+ ≤ft(1+x²))+c, where c is a constant of integration.

Correct Answer is : 2xtan⁡−1x−log⁡(1+x2)+c2 x ^{-1} x- ≤ft(1+x²)+c2xtan−1x−log(1+x2)+c, where c is a constant of integration.

Test Index

MHTCET 10 May 2024 Shift 2 Paper

Question : 146 of 150

Marks: +1, -0

\int \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x=

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

Solution:

Concept:

Use the substitution

x = \tan t

to simplify the inverse trigonometric expression, then integrate by parts.

Explanation:

Let

x = \tan t

, so

dx = \sec^2 t \, dt

Then

\frac{2x}{1+x^2} = \frac{2\tan t}{1+\tan^2 t} = \frac{2\tan t}{\sec^2 t} = 2\sin t \cos t = \sin 2t

So the integral becomes

I = \int \sin^{-1}(\sin 2t) \sec^2 t \, dt = \int 2t \sec^2 t \, dt

Integrating by parts:

I = 2\left[t \tan t - \int \tan t \, dt\right] = 2t \tan t + 2\ln|\cos t| + c

Since

t = \tan^{-1} x

, we have

\tan t = x

and

\cos t = \frac{1}{\sqrt{1+x^2}}

Thus

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

Answer:

2x \tan^{-1}x - \ln(1+x^2) + c

, which corresponds to option A.

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