What is the antiderivative of #(x^3)sqrt(x^2 + 1)) dx#?

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Answer 1 · 2015-04-13T17:10:08

The answer is: #1/15(x^2+1)sqrt(x^2+1)(3x^2-2)+c#.

If we make the substitution:

#x=sinhtrArrdx=coshtdt#, than:

#intx^3sqrt(x^2+1)dx=intsinh^3tsqrt(sinh^2t+1)*coshtdt=#

#=intsinh^3tsqrt(cosh^2t)*coshtdt=intsinh^3tcosh^2tdt=#

#=intsinh^2tsinhtcosh^2tdt=int(cosh^2t-1)sinhtcosh^2tdt=#

#=int(sinhtcosh^4t-sinhtcosh^2t)dt=#

#=cosh^5t/5-cosh^3t/3+c=(1)#

And, since #cosht=sqrt(sinh^2t+1)=sqrt(x^2+1)#,

#(1)=(sqrt(x^2+1))^5/5-(sqrt(x^2+1))^3/3+c=#

#=(x^2+1)^2sqrt(x^2+1)/5-(x^2+1)sqrt(x^2+1)/3+c=#

#=(x^2+1)sqrt(x^2+1)[(x^2+1)/5-1/3]+c=#

#=(x^2+1)sqrt(x^2+1)((3x^2+3-5)/15)+c=#

#=1/15(x^2+1)sqrt(x^2+1)(3x^2-2)+c#.