Can someone show me the steps to finding the indefinite integral/antiderivative of:

x^2 / sqrt(x^2+9)

Let y=x/3
dy = dx/3
3dy = dx

x²dx / sqrt(x² + 9) = 27y²dy / sqrt(9y² + 9) = 27y²dy / [3sqrt(y²+1)] = 9y²dy / sqrt(y²+1)

Now, let y=sinh(t)
dy = cosh(t)dt

x²dx / sqrt(x²+9) = 9sinh²(t)cosh(t)dt / sqrt(sinh²(t)+1) = 9sinh²(t)cosh(t)dt / cosh(t) = 9sinh²(t)dt

And since sinh²(t) = ( cosh(2t) - 1 ) / 2 :

x²dx / sqrt(x²+9) = 9(cosh(2t) - 1)dt / 2

I guess you can conclude at this point.

This post was edited by feanur on Nov 17 2013 04:58pm