Simplifying Integration by Parts: Solving ∫ln(x+x^2)dx Using the Hint x(1+x)

[klancello]

Hello. I'm attempting to integrat ∫ln(x+x^2)dx
Our professor gave us the hint of x(1+x)
I believe u= ln(x+x^2) and du=1+2x/x+x^2
I am not sure what dv should be
Any help would be greatly appreciated! Thanks

 

[epenguin]

Hello. I'm attempting to integrat ∫ln(x+x^2)dx
Our professor gave us the hint of x(1+x)
I believe u= ln(x+x^2) and du=1+2x/x+x^2
I am not sure what dv should be
Any help would be greatly appreciated! Thanks

I would break your expression up a bit first into simpler things. What is the log of a product?

Though there may be more than one way.

 

[david sh]

i think this answer is:
1/4x(-x+2(x+2)lnx-4)+C

 

[david sh]

u=ln(x^2 + 1)-----)du=2x/(x^2+1)
v=x--------------)dv= dx.

note: der( u*v) = u dv+ v du.
and then Integrate both sides and you get :
uv= int(u dv)+ int (v du). Switch it around and you get int(u dv) =uv- int (v du)
So the integral is ln(x^2 + 1)*x- int(2x^2/x^2+1)
Next, integrate the last part. int(2x^2/(x^2+1)). Take the 2 out for a minute. Then add a one and subtract a one in the top. int(x^2+1-1)/(x^2+1). Separate them and the integral is int(1-1/(x^2+1)). That's x - tan inverse x. Put the 2 back in, and it's 2x-2 taninvers x.
Put them together and you will have ln(x^2 + 1)*x-2x+2 taninverse x+C.

 

[HallsofIvy]

Ouch! Where did $ln(x^2+ 1)$ come from? That is much harder than the given $ln(x^2+ x)$

What epenguin was suggesting was that you use the fact that $ln(x+ x^2)= ln(x(1+ x))= ln(x)+ ln(1+ x)$.