{x^{4} ≤ft( ^{-1} x^{5})}{1+x¹0}} d x equals

Correct Answer is : sin⁡(tan⁡−1x5)5+c { ≤ft( ^{-1} x^{5})}{5}+c5sin(tan−1x5)+c, where c is the constant of integration

{ ≤ft( ^{-1} x^{5})}{5}+c, where c is the constant of integration

x^{4} ≤ft( ^{-1} x^{5})+c, where c is the constant of integration

{ ≤ft( ^{-1} x^{5})}{4}+c, where c is the constant of integration

≤ft( ^{-1} x^{5})+c, where c is the constant of integration

Correct Answer is : sin⁡(tan⁡−1x5)5+c { ≤ft( ^{-1} x^{5})}{5}+c5sin(tan−1x5)+c, where c is the constant of integration

Test Index

MHTCET 25 Apr 2025 Shift 2 Paper

Question : 138 of 150

Marks: +1, -0

\int \frac{x^{4} \cos \left(\tan ^{-1} x^{5}\right)}{1+x^{10}} d x

equals

$\quad \frac{\sin \left(\tan ^{-1} x^{5}\right)}{5}+c$ , where c is the constant of integration
$\quad x^{4} \sin \left(\tan ^{-1} x^{5}\right)+c$ , where c is the constant of integration
$\frac{\sin \left(\tan ^{-1} x^{5}\right)}{4}+c$ , where c is the constant of integration
$\quad \cos \left(\tan ^{-1} x^{5}\right)+c$ , where c is the constant of integration

Solution:

Concept:

Use substitution to simplify the integral into a standard form of

\sin

Explanation:

Let

I = \int \frac{x^{4} \cos \left(\tan ^{-1} x^{5}\right)}{1+x^{10}} \, dx

Put

t = \tan^{-1}\left(x^{5}\right)

Then,

\frac{1}{1+\left(x^{5}\right)^2} \cdot 5x^{4} \, dx = dt

So,

\frac{5x^{4} \, dx}{1+x^{10}} = dt

Hence,

I = \frac{1}{5} \int \cos t \, dt

This gives

I = \frac{1}{5} \sin t + c

Substitute back

t = \tan^{-1}\left(x^{5}\right)

I = \frac{1}{5} \sin \left(\tan^{-1} x^{5}\right) + c

Answer:

\frac{\sin \left(\tan ^{-1} x^{5}\right)}{5}+c

, where

c

is the constant of integration.

Therefore, the correct option is A.

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