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

Question

1621 views around the world

You can reuse this answer
Creative Commons License

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 $cosht=sqrt(sinh^2t+1)=sqrt(x^2+1)$ ,

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

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

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

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

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

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