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

Category: QuestionsFind \int \,7x^8-4e^{2x}-\frac{2}{x^2} \,dx.
Editor">Editor Staff asked 11 months ago

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
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1 Answers
Editor">Editor Staff answered 11 months ago

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