# Integrate (x^2+9) e^{2x} dx

Category: QuestionsIntegrate (x^2+9) e^{2x} dx
Editor">Editor Staff asked 11 months ago

Integrate (x^2+9) e^{2x} dx

(a) \frac{e^{x}}{2} (x^2+x-\frac{39}{4})+C

(b) \frac{e^{2x}}{2} (x^2+x-\frac{35}{4})+C

(c) \frac{e^{2x}}{2} (x^2+x-\frac{48}{4})+C

(d) \frac{e^{x}}{2} (x^2+x-\frac{25}{4})+C

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Editor">Editor Staff answered 11 months ago

Correct answer is (b) \frac{e^{2x}}{2} (x^2+x-\frac{35}{4})+C

For explanation: By using ∫ u.v dx=u∫ v dx-∫ u'(∫ v dx), we get

\int (x^2+9) \,e^x \,dx=(x^2+9) \int e^{2x} \,dx-\int (x^2+9)’\int e^{2x} \,dx

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

Again, applying integration by parts for integrating \int \,x \,e^{2x} \,dx

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

=\frac{xe^{2x}}{2}-\int \frac{e^{2x}}{2} \,dx

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

∴\int \,(x^2+9) \,e^x \,dx=\frac{(x^2+9) e^{2x}}{2}+\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}

=\frac{e^{2x}}{2}(x^2+9+x-\frac{1}{4})

=\frac{e^{2x}}{2}(x^2+x-\frac{35}{4})+C.