{e^{ ^{-1} x}}{1+x²}≤ft[≤ft( ^{-1} {1+x²})²+ ^{-1}≤ft({1-x²}{1+x²})] d x, where x > 0 is

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

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

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

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

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

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

Test Index

MHTCET 16 May 2024 Shift 2 Paper

Question : 138 of 150

Marks: +1, -0

\int \frac{e^{\tan ^{-1} x}}{1+x^{2}}\left[\left(\sec ^{-1} \sqrt{1+x^{2}}\right)^{2}+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right] d x

x > 0

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

Solution:

Concept:

The integral simplifies by substituting

x = \tan t

, which converts the inverse trigonometric expressions into simple algebraic forms.

Explanation:

Let

I = \int \frac{e^{\tan^{-1}x}}{1+x^2}\left[\left(\sec^{-1}\sqrt{1+x^2}\right)^2 + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]dx

Put

x = \tan t

, so

t = \tan^{-1}x

Then

dx = \sec^2 t \, dt

and

1 + x^2 = \sec^2 t

Also,

\sqrt{1+x^2} = \sec t

, and since

x > 0

, we have

\sec^{-1}(\sec t) = t

And

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

Thus

\cos^{-1}(\cos 2t) = 2t

The integral becomes

I = \int e^t \left[t^2 + 2t\right]dt

Using the property

\int e^t\left[f(t) + f'(t)\right]dt = e^t f(t) + c

with

f(t) = t^2

, we get

I = e^t t^2 + c

Substituting back

t = \tan^{-1}x

gives

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

Answer:

Option B:

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

, where

c

is a constant of integration.

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