How do I find the integral intt^2/(t+4)dt ?

1 Answer
Sep 30, 2014

=t^2/2-4t-24+16ln(t+4)+c, where c is a constant

Explanation :

=intt^2/(t+4)dt

let's t+4=u, then dt=du

=int(u-4)^2/udu

=int(u^2-8u+16)/udu

=int(u^2/u-8u/u+16/u)du

=int(u-8+16/u)du

=intudu-int8du+int16/udu

=u^2/2-8u+16lnu+c, where c is a constant

Substituting u back yields,

=(t+4)^2/2-8(t+4)+16ln(t+4)+c, where c is a constant

Simplifying further,

=t^2/2+8+4t-8t-32+16ln(t+4)+c, where c is a constant

=t^2/2-4t-24+16ln(t+4)+c, where c is a constant