Question

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 1

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

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

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

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

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