How to integrate #int(x^3 dx)/sqrt( x^2 + 4)# by using substitution?

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Answer 1 · 2018-04-21T11:26:01

#I=1/3sqrt(x^2+4) (x^2-8)+c#

Here,

#I=intx^3/sqrt(x^2+4)dx=intx^2/sqrt(x^2+4)xdx#

Take,

#sqrt(x^2+4)=u=>x^2+4=u^2=>x^2=u^2-4#

So,

#:.2xdx=2udu=>xdx=udu#

#I=int(u^2-4)/uxxudu#

#=int(u^2-4)du#

#=u^3/3-4u+c#

#=u/3(u^2-12)+c,where,u=sqrt(x^2+4)#

#=1/3sqrt(x^2+4)(x^2+4-12)+c#

#I=1/3sqrt(x^2+4) (x^2-8)+c#