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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»

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

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

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

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

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


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