Indefinite Integration – Integration by Parts – JEE Main 23 Jan 2026 Shift 1

Solution:

We are given the integral:
$$ I = \int e^{x} \left[ \frac{2-x^{2}}{\sqrt{1+x}(1-x)^{3/2}} \right] dx $$

Rearrange the numerator $2-x^2$ as $1 + (1-x^2)$:
$$ I = \int e^{x} \left[ \frac{1+(1-x^2)}{(1+x)^{1/2}(1-x)^{3/2}} \right] dx $$

Split the fraction into two parts:
$$ I = \int e^{x} \left[ \frac{1-x^2}{(1+x)^{1/2}(1-x)^{3/2}} + \frac{1}{(1+x)^{1/2}(1-x)^{3/2}} \right] dx $$

Simplify the first term:
$$ \frac{(1-x)(1+x)}{(1+x)^{1/2}(1-x)^{3/2}} = \frac{(1+x)^{1/2}}{(1-x)^{1/2}} = \sqrt{\frac{1+x}{1-x}} $$

Now the integral becomes:
$$ I = \int e^{x} \left[ \sqrt{\frac{1+x}{1-x}} + \frac{1}{\sqrt{1+x}(1-x)^{3/2}} \right] dx $$

This is of the form $\int e^x (g(x) + g'(x)) dx = e^x g(x) + C$.
Let’s verify the derivative of $g(x) = \sqrt{\frac{1+x}{1-x}}$:
$$ g'(x) = \frac{d}{dx}\left(\frac{1+x}{1-x}\right)^{1/2} = \frac{1}{2}\left(\frac{1+x}{1-x}\right)^{-1/2} \cdot \frac{(1-x)(1) – (1+x)(-1)}{(1-x)^2} $$

$$ g'(x) = \frac{1}{2}\sqrt{\frac{1-x}{1+x}} \cdot \frac{2}{(1-x)^2} = \frac{1}{\sqrt{1+x}(1-x)^{3/2}} $$

Thus, the integral is:
$$ f(x) = e^x \sqrt{\frac{1+x}{1-x}} + C $$

Given $f(0) = 0$:
$$ e^0 \sqrt{\frac{1+0}{1-0}} + C = 0 \Rightarrow 1 + C = 0 \Rightarrow C = -1 $$

Now find $f(1/2)$:
$$ f\left(\frac{1}{2}\right) = e^{1/2} \sqrt{\frac{1+1/2}{1-1/2}} – 1 = \sqrt{e} \sqrt{\frac{3/2}{1/2}} – 1 $$

$$ = \sqrt{3e} – 1 $$

Ans. (1)