Key to "Find the indefinite integral. ∫xe2x(2x+1)2dx"

vazelinahS

Answered question

2021-11-09

Find the indefinite integral.

\int \frac{x e^{2 x}}{{(2 x + 1)}^{2}} d x

Adnaan Franks

Skilled2021-11-10Added 92 answers

Step 1
Given Integral,

\int \frac{x e^{2 x}}{{(2 x + 1)}^{2}} d x

We use the integration by parts rule to solve this integration.

\int u v^{'} = u v - \int v d u

...(A)
Step 2
here,

u = x e^{2 x} and v^{'} = \frac{1}{{(2 x + 1)}^{2}}

\frac{d u}{d x} = x (2 e^{2 x}) + e^{2 x} (1) = e^{2 x} (2 x + 1)

d u = e^{2 x} (2 x + 1)

...(1)

v^{'} = \frac{1}{{(2 x + 1)}^{2}}

v = \frac{1}{2 (2 x + 1)}

...(2)
Step 3
plug values of (1) and (2) in (A),

\int \frac{x e^{2 x}}{{(2 x + 1)}^{2}} d x = - \frac{e^{2 x}}{2 (2 x + 1)} - \int - \frac{1}{2} e^{2 x} d x

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

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

So,

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

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?