How do I find the integral t2t+4dt ?

1 Answer
Sep 30, 2014

=t224t24+16ln(t+4)+c, where c is a constant

Explanation :

=t2t+4dt

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

=(u4)2udu

=u28u+16udu

=(u2u8uu+16u)du

=(u8+16u)du

=udu8du+16udu

=u228u+16lnu+c, where c is a constant

Substituting u back yields,

=(t+4)228(t+4)+16ln(t+4)+c, where c is a constant

Simplifying further,

=t22+8+4t8t32+16ln(t+4)+c, where c is a constant

=t224t24+16ln(t+4)+c, where c is a constant