JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Indefinite Integrals

Factor the denominator as $x^4+x=x(x^3+1)$ and split by partial fractions. The constant piece is $\frac{-1}{x}$, since at $x=0$ the numerator over $(x^3+1)$ is $-1$; the remainder is $\frac{(2x^3-1)+(x^3+1)}{x(x^3+1)}=\frac{3x^3}{x(x^3+1)}=\frac{3x^2}{x^3+1}$. So the integrand is $-\frac{1}{x}+\frac{3x^2}{x^3+1}$, and integrating term by term gives $-\ln |x|+\ln |x^3+1|=\ln \left|\frac{x^3+1}{x}\right|+C$. The division inside the logarithm is by $x$, not by $x^2$.