Question

Find the integral of \(\frac{4x^4-3x^2}{x^3}\).

(a) 7x^2-3 log⁡x^3+C

(b) 2x^2-3 log⁡x+C

(c) x^2-log⁡x+C

(d) 2x^2+3 log⁡x+C

I got this question by my college director while I was bunking the class.

This key question is from Integration as an Inverse Process of Differentiation topic in section Integrals of Mathematics – Class 12

Answer 1

The correct choice is (b) 2x^2-3 log⁡x+C

Easiest explanation: To find \(\int \frac{4x^4-3x^2}{x^3} dx\)

\(\int \frac{4x^4-3x^2}{x^3} \,dx=\int \frac{4x^4}{x^3} – \frac{3x^2}{x^3} \,dx\)

\(\int \frac{4x^4-3x^2}{x^3} \,dx=\int 4x dx-\int \frac{3}{x} dx\)

\(\int \frac{4x^4-3x^2}{x^3} \,dx=\frac{4x^2}{2}-3 log⁡x\)

∴ \(\int \frac{4x^4-3x^2}{x^3} \,dx=2x^2-3 \,log⁡x+C\).