Question

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

Answer 1

`=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