How do I integrate by substitution?

#int(1-x)/(1+x)^3dx#

3 Answers
Parabola
Apr 20, 2018

#(x)/((x+1)^2)+C#

Explanation:

We let #x+1=u#.

Then:

#du=1dx#

Substitute:

#=>int(2-u)/(u^3)du#

#=>int(2)/(u^3)-u/(u^3)du#

#=>int(2)/(u^3)du-intu/(u^3)du#

#=>2intu^-3du-intu^-2du#

Remember that

#intx^ndx=x^(n+1)/(n+1)#

#=>2*u^-2/-2-u^-1/-1#

#=>-u^-2+u^-1#

#=>1/u-1/u^2#
#=>(u-1)/(u^2)# substitute.

#=>((x+1)-1)/((x+1)^2)#

#=>(x)/((x+1)^2)+C#
That is the answer!

Ultrilliam
Apr 20, 2018

Use #z = 1+ x#

Explanation:

#int \ (1-x)/(1+x)^3 \ dx#

Well the denominator is not helpful so why not try:

#z = 1+ x# ie #x = z-1#

That gives you:

#int \ (1-z + 1)/z^3 \ d(z-1)#

#= int \ (2-z)/z^3 \ dz#

#= \ -1/z^(2) + 1/z + C#

#= \- 1/(1+x)^2 + 1/(1+x) + C#

#= \ x/(1+x)^2 + C#

Ratnaker Mehta
Apr 20, 2018

# x/(x+1)^2+C#.

Explanation:

If the substitution is not must, then,

#I=int(1-x)/(1+x)^3dx=-int(x-1)/(x+1)^3dx#,

#=-int{(x+1)-2}/(x+1)^3dx#,

#=-int{(x+1)/(x+1)^3-2/(x+1)^3}dx#,

#=-int1/(x+1)^2dx+2int1/(x+1)^3dx#,

#=-(x+1)^(-2+1)/(-2+1)+2*(x+1)^(-3+1)/(-3+1)#.

#=1/(x+1)-1/(x+1)^2#,

#={(x+1)-1}/(x+1)^2#.

# rArr I=x/(x+1)^2+C#.

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