How do you find the integral int (x+1)/(x^2+2x+2)^3dxx+1(x2+2x+2)3dx using substitution?

2 Answers
Dec 15, 2016

The answer is =-1/(4(x^2+2x+2))+C=14(x2+2x+2)+C

Explanation:

We use intx^ndx=x^(n+1)/(n+1)+C (n!=0)xndx=xn+1n+1+C(n0)

Let u=x^2+2x+2u=x2+2x+2

du=(2x+2)dxdu=(2x+2)dx

(x+1)dx=(du)/2(x+1)dx=du2

int((x+1)dx)/(x^2+2x+2)^3=1/2int(du)/(u^3)=1/2intu^(-3)du(x+1)dx(x2+2x+2)3=12duu3=12u3du

=1/2*u^(-3+1)/(-3+1)=12u3+13+1

=1/2*u^(-2)/(-2)=12u22

=-1/(4u^2)=14u2

=-1/(4(x^2+2x+2))+C=14(x2+2x+2)+C

Dec 15, 2016

-1/(4(x^2+2x+2)^2)+C14(x2+2x+2)2+C

Explanation:

You first decide which part of one looks like the other. If you make
U= x^2+2x+2U=x2+2x+2
dU= 2x+2dU=2x+2
Therefore, all you need to do is multiply the top by 2 and you have the same thing as your dU.
You then have multiply the inside by 2 but also the outside by 1/2.
Once you have this, you can then substitute your U and dU in your situation.
1/2int_ . 1/((U)^3) du12.1(U)3du

you final recieve after your done
1/2(-1/(2u^2))12(12u2)

Then you finally substitute your U back in the situation and you have

-1/(4(x^2+2x+2)^2 )+ C14(x2+2x+2)2+C