JEE Main 2026 Jan 23 Shift 1, Mathematics Q2: Integration as an Inverse Process of Differentiation

Re-write the integrand. Split $2-x^2=(1-x^2)+1$:

$f(x)=\int e^x\left(\frac{1-x^2}{\sqrt{1+x}(1-x{)}^{3/2}}+\frac{1}{\sqrt{1+x}(1-x{)}^{3/2}}\right)dx$

The first term simplifies because $1-x^2=(1+x)(1-x)$:

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

Recognise an exact derivative. Let $g(x)=e^x\sqrt{\frac{1+x}{1-x}}$. Then $g^{\prime }(x)=e^x\sqrt{\frac{1+x}{1-x}}+e^x\cdot \frac{1}{\sqrt{1+x}(1-x{)}^{3/2}}$, which matches the integrand.

$\therefore f(x)=e^x\sqrt{\frac{1+x}{1-x}}+C$. Using $f(0)=0$ gives $C=-1$.

$f\left(\frac{1}{2}\right)=e^{1/2}\sqrt{\frac{3/2}{1/2}}-1=\sqrt{e}\cdot \sqrt{3}-1=\sqrt{3e}-1$.