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Integrate xe^2x.(a) \(\frac{e^{2x}}{4} (x-\frac{1}{4})+C\)(b) \(\frac{e^{2x}}{4} (2x-1)+C\)(c) \(\frac{e^{2x}}{2} (2x-1)+C\)(d) \(\frac{e^{2x}}{4} (x+1)+C\)This question was posed to me in class test.I'd like to ask this question from Integration by Parts in division Integrals of Mathematics – Class 12

Answer»

The correct option is (b) \(\frac{e^{2x}}{4} (2x-1)+C\)

Explanation: By using the FORMULA \(\INT u.V \,dx=u\int \,v \,dx-\int u'(\int \,v \,dx)\) we GET

\(\int XE^{2x} dx=x\int e^{2x} dx-\int (x)’\int e^{2x} \,dx\)

=\(\frac{xe^{2x}}{2}-\int 1.\frac{e^{2x}}{2} \,dx\)

=\(\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\)

=\(\frac{e^{2x}}{4} (2x-1)+C\)



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