# Thread: Calculus - : ∫(x^2)/sqrt(x^2+1)

1. ## Calculus - : ∫(x^2)/sqrt(x^2+1)

Question: ∫(x^2)/sqrt(x^2+1)
u=x^2+1 , x^2= u-1, du=2xdx
∫(u-1)/sqrt(u) , expand ∫u/sqrt(u) - 1/sqrt(u)
Integrate: 2/3(u^(3/2)) - 2u^(1/2) + c
My answer: [ 2/3(x^2+1)^(3/2) - 2(x^2+1) + c ]

when I derived the expression I got the answer I got was not the same as the integrand. Help? Thank you !

2. Originally Posted by lecheflan
Question: ∫(x^2)/sqrt(x^2+1) $\ \ \ \ \ \ \$You're missing the "dx." And you have to account for it when you substitute.

u=x^2+1 , x^2= u-1, du=2xdx

∫(u-1)/sqrt(u) , expand ∫u/sqrt(u) - 1/sqrt(u) $\ \ \ \ \ \ \$So, you're missing the corresponding "du." That changes the integrand significantly.

when I derived the expression I got the answer I got was not the same as the integrand. Help? Thank you !
du = 2xdx

However, when I looked up the answer in an Internet integration calculator, it involves the inverse hyperbolic sine function.

That puts this problem further along integration techniques and difficulty, unless you're allowed to use a table of integration, for instance.