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

Sep 30, 2014

$= {t}^{2} / 2 - 4 t - 24 + 16 \ln \left(t + 4\right) + c$, where $c$ is a constant

Explanation :

$= \int {t}^{2} / \left(t + 4\right) \mathrm{dt}$

let's $t + 4 = u$, then $\mathrm{dt} = \mathrm{du}$

$= \int {\left(u - 4\right)}^{2} / u \mathrm{du}$

$= \int \frac{{u}^{2} - 8 u + 16}{u} \mathrm{du}$

$= \int \left({u}^{2} / u - 8 \frac{u}{u} + \frac{16}{u}\right) \mathrm{du}$

$= \int \left(u - 8 + \frac{16}{u}\right) \mathrm{du}$

$= \int u \mathrm{du} - \int 8 \mathrm{du} + \int \frac{16}{u} \mathrm{du}$

$= {u}^{2} / 2 - 8 u + 16 \ln u + c$, where $c$ is a constant

Substituting $u$ back yields,

$= {\left(t + 4\right)}^{2} / 2 - 8 \left(t + 4\right) + 16 \ln \left(t + 4\right) + c$, where $c$ is a constant

Simplifying further,

$= {t}^{2} / 2 + 8 + 4 t - 8 t - 32 + 16 \ln \left(t + 4\right) + c$, where $c$ is a constant

$= {t}^{2} / 2 - 4 t - 24 + 16 \ln \left(t + 4\right) + c$, where $c$ is a constant