Question

Find \(\int \,7x^8-4e^{2x}-\frac{2}{x^2} \,dx\).

(a) \(\frac{7x^4}{4}-2e^{2x}+\frac{2}{x}+C\)

(b) \(\frac{7x^4}{4}+2e^{2x}+\frac{2}{x}+C\)

(c) \(\frac{7x^4}{4}-2e^{2x} \frac{2}{x^2}+C\)

(d) \(\frac{7x^4}{8}+2e^{2x}-\frac{4}{x}+C\)

The question was posed to me in semester exam.

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

Answer 1

Correct option is (a) \(\frac{7x^4}{4}-2e^{2x}+\frac{2}{x}+C\)

The best I can explain: To find:\(\int 7x^8-4e^{2x}-\frac{2}{x^2} dx\)

\(\int \,7x^8-4e^{2x}-\frac{2}{x^2} \,dx=\int 7x^9 dx-4\int e^{2x} dx-2\int \frac{1}{x}^2 dx\)

\(\int \,7x^8-4e^{2x}-\frac{2}{x^2} \,dx=\frac{7x^{9+1}}{9+1}-\frac{4e^{2x}}{2}-\frac{2x^{-2+1}}{-2+1}\)

∴\(\int \,7x^8-4e^{2x}-\frac{2}{x^2} dx=\frac{7x^{10}}{10}-2e^{2x}+\frac{2}{x}+C\)