Question

integrate : xe2x / (1+2x) 2

Answer 1

Another method:

I=∫xe2x(2x+1)2dx

We can try integration by parts with u=e2xand dv=x(2x+1)2dx.

Note that v=∫x(2x+1)2dx. Letting t=2x+1, this implies that x=12(t−1) and that dt=2dx⇒dx=12dt, so v=∫12(t−1)t212dt=14∫(1t−1t2)dt=14ln|t|+14t...

Also, du=2e2xdx.

Then:

I=14e2x(ln|2x+1|+12x+1)−∫2e2x14(ln|2x+1|+12x+1)dx

I=e2x4ln|2x+1|+e2x4(2x+1)−12∫e2xln|2x+1|dx−12∫e2x2x+1dx

Trying IBP on the second integral, let:

u=−12e2x⇒du=−e2xdx
dv=dx2x+1⇒v=12ln|2x+1|

So:

I=e2x4ln|2x+1|+e2x4(2x+1)−12∫e2xln|2x+1|dx−e2x4ln|2x+1|+12∫e2xln|2x+1|dx

I=e2x4(2x+1)+C

Answer 2

Another method:

I=∫xe2x(2x+1)2dx

We can try integration by parts with u=e2x and dv=x(2x+1)2dx.

Note that v=∫x(2x+1)2dx. Letting t=2x+1, this implies that x=12(t−1) and that dt=2dx⇒dx=12dt, so v=∫12(t−1)t212dt=14∫(1t−1t2)dt=14ln|t|+14t...

Also, du=2e2xdx.

Then:

I=14e2x(ln|2x+1|+12x+1)−∫2e2x14(ln|2x+1|+12x+1)dx

I=e2x4ln|2x+1|+e2x4(2x+1)−12∫e2xln|2x+1|dx−12∫e2x2x+1dx

Trying IBP on the second integral, let:

u=−12e2x⇒du=−e2xdx
dv=dx2x+1⇒v=12ln|2x+1|

So:

I=e2x4ln|2x+1|+e2x4(2x+1)−12∫e2xln|2x+1|dx−e2x4ln|2x+1|+12∫e2xln|2x+1|dx

I=e2x4(2x+1)+C