{e^{ ^{-1} x}}{≤ft(1+x²)}≤ft(1+x+x²) d x=

Correct Answer is : xetan⁡−1x+cx e^{ ^{-1} x}+cxetan−1x+c

{x e^{ ^{-1} x}}{≤ft(1+x²)}+c

Correct Answer is : xetan⁡−1x+cx e^{ ^{-1} x}+cxetan−1x+c

Test Index

COMEDK 2025 Afternoon Solved Paper

Question : 152 of 180

Marks: +1, -0

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

Solution:

Concept:

Use substitution and recognize the integrand as the derivative of a product.

Explanation:

Let

u = \tan^{-1} x

Then

du = \frac{dx}{1+x^2}

and

x = \tan u

Rewrite the integral:

\displaystyle \int \frac{e^{\tan^{-1}x}}{1+x^2}(1+x+x^2)\,dx = \int e^{u}\,(1+\tan u + \tan^2 u)\,du

Simplify using

\tan^2 u = \sec^2 u - 1

1 + \tan u + \tan^2 u = 1 + \tan u + (\sec^2 u - 1) = \tan u + \sec^2 u

So the integral becomes

\displaystyle \int e^{u}(\tan u + \sec^2 u)\,du

Notice that

\displaystyle \frac{d}{du}\left(e^{u}\tan u\right) = e^{u}\tan u + e^{u}\sec^2 u = e^{u}(\tan u + \sec^2 u)

Thus the integrand matches the derivative of

e^{u}\tan u

Therefore

\displaystyle \int e^{u}(\tan u + \sec^2 u)\,du = e^{u}\tan u + C

Substitute back

u = \tan^{-1}x

and

\tan u = x

e^{\tan^{-1}x} \cdot x + C = x e^{\tan^{-1}x} + C

Answer:

Option B:

x e^{\tan^{-1} x} + c

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