Re: Using integral by parts

Integration by Parts is not the best method to use here.

$\displaystyle \displaystyle \begin{align*} \int{\frac{t+5}{\sqrt{8-t}}\,dt} = -\int{-\frac{t+5}{\sqrt{8-t}}\,dt} \end{align*}$

Let $\displaystyle \displaystyle \begin{align*} u = 8 - t \implies du = -dt \end{align*}$ and note that $\displaystyle \displaystyle \begin{align*} t + 5 = 13 - u \end{align*}$ and the integral becomes

$\displaystyle \displaystyle \begin{align*} -\int{-\frac{t + 5}{\sqrt{8-t}}\,dt} &= -\int{\frac{13 - u}{\sqrt{u}}\,du} \\ &= -\int{13u^{-\frac{1}{2}} - u^{\frac{1}{2}} \, du} \end{align*}$

This should now be easy to solve.

Re: Using integral by parts

Quote:

Originally Posted by

**Prove It** Integration by Parts is not the best method to use here.

$\displaystyle \displaystyle \begin{align*} \int{\frac{t+5}{\sqrt{8-t}}\,dt} = -\int{-\frac{t+5}{\sqrt{8-t}}\,dt} \end{align*}$

Let $\displaystyle \displaystyle \begin{align*} u = 8 - t \implies du = -dt \end{align*}$ and note that $\displaystyle \displaystyle \begin{align*} t + 5 = 13 - u \end{align*}$ and the integral becomes

$\displaystyle \displaystyle \begin{align*} -\int{-\frac{t + 5}{\sqrt{8-t}}\,dt} &= -\int{\frac{13 - u}{\sqrt{u}}\,du} \\ &= -\int{13u^{-\frac{1}{2}} - u^{\frac{1}{2}} \, du} \end{align*}$

This should now be easy to solve.

Unfortunately I have to use integration by parts as instructed by my teacher but this should help me come up with the answer. Thank you.