How do you integrate #int (x^2+2x)/(x^2+2x+1)# using substitution?

Question

8811 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-13T08:04:28

The answer is #=(x+1)+1/(x+1)+C#

Let's factorise,

Therefore,

#int((x^2+2x)dx)/(x^2+2x+1)#

#=int(x(x+2)dx)/(x+1)^2#

Let #u=x+1#

#du=dx#

#x+2=u+1#

#x=u-1#

So,

#int(x(x+2)dx)/(x+1)^2=int((u+1)(u-1)du)/(u^2)#

#=int((u^2-1)du)/u^2#

#=int(1-1/u^2)du#

#=u+1/u#

#=(x+1)+1/(x+1)+C#

Answer 2 · 2017-02-13T17:08:09

I have found the following results with one substitution:
enter image source here