# Calculus - Integrating

posted by Liam

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 took the derivative of this to check my answer, it was not (x^2)/sqrt(x^2+1). Please help :/

1. Steve

The problem is that you did not have du in your integrals. You lacked the extra 2x dx in your integrands.

u = x^2+1
du = 2x dx
x = √(u-1)

So you have

∫x/√(x^2+1) x dx
= 1/2 ∫√(u-1)/u du
which isn't really much better

Try using a trig substitution:
x = tanθ
x^2+1 = tan^2θ + 1 = sec^2θ
dx = sec^2θ dθ

∫x^2/√(x^2+1) dx
= ∫(sec^2θ-1)/secθ sec^2θ dθ
= ∫sec^3θ-secθ dθ
= 1/2 secθ tanθ - 1/2 log tan(π/4 - θ/2)
= 1/2 x√(x^2+1) - 1/2 log (√(1+x^2) - x)
= 1/2 (x√(x^2+1) - sinh-1x)

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