How do I integrate by substitution?
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"How do I integrate by substitution?"
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!
#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#
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#.