Q61 / 180 · COMEDK 2025 (10 May Shift 2) · Mathematics
$\int \dfrac{e^{\tan ^{-1} x}}{\left(1+x^2\right)}\left(1+x+x^2\right) d x=$
- A$\dfrac{x e^{\tan ^{-1} x}}{\left(1+x^2\right)}+c$
- B$e^{\tan ^{-1} x}+c$
- Correct answerC$x e^{\tan ^{-1} x}+c$
- D$\dfrac{e^{\tan ^{-1} x}}{\left(1+x^2\right)}+c$
Solution
Let $I = \int \dfrac{e^{\tan^{-1} x}}{1+x^2} (1+x+x^2) dx$.
Substitute $t = \tan^{-1} x$, so $dt = \dfrac{1}{1+x^2} dx$.
Then $x = \tan t$. The integral becomes $I = \int e^t (1 + \tan t + \tan^2 t) dt$.
Using the identity $1 + \tan^2 t = \sec^2 t$, we have $I = \int e^t (\sec^2 t + \tan t) dt$.
Recall the standard integral form $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$.
Here, let $f(t) = \tan t$, then $f'(t) = \sec^2 t$.
Thus, $I = e^t \tan t + c$.
Substituting back $t = \tan^{-1} x$ and $\tan t = x$, we get $I = e^{\tan^{-1} x} \cdot x + c$.
Answer: $x e^{\tan ^{-1} x}+c$
Substitute $t = \tan^{-1} x$, so $dt = \dfrac{1}{1+x^2} dx$.
Then $x = \tan t$. The integral becomes $I = \int e^t (1 + \tan t + \tan^2 t) dt$.
Using the identity $1 + \tan^2 t = \sec^2 t$, we have $I = \int e^t (\sec^2 t + \tan t) dt$.
Recall the standard integral form $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$.
Here, let $f(t) = \tan t$, then $f'(t) = \sec^2 t$.
Thus, $I = e^t \tan t + c$.
Substituting back $t = \tan^{-1} x$ and $\tan t = x$, we get $I = e^{\tan^{-1} x} \cdot x + c$.
Answer: $x e^{\tan ^{-1} x}+c$
Hence, the correct answer is option: 'C'.