What is the integral of #int x^3/(1+x^2)dx#?

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Answer 1 · 2018-03-21T15:51:00

The answer is #=1/2(1+x^2)-1/2ln(1+x^2)+C#

Perform the substitution

#u=1+x^2#, #=>#, #du=2xdx#

#=>#, #x^2=u-1#

Therefore,

#int(x^3dx)/(1+x^2)=int(x^2xdx)/(1+x^2)=1/2int((u-1)du)/(u)#

#=1/2int1du-1/2int(du)/u#

#=1/2u-1/2lnu#

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

Answer 2 · 2018-03-21T16:02:04

#int x^3/(x^2+1)*dx=x^2/2-1/2Ln(x^2+1)+C#

#int x^3/(x^2+1)*dx#

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

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

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

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