# Using integral by parts

• Feb 3rd 2013, 06:56 PM
kuppina
Using integral by parts
Hello everyone it's me again...

For this question I have to use integral by parts to find the integral of (t+5)/(sqrt 8-t)

I have:
u= t+5
v' = (8-t)^-1/2

u'= 1
v= 2(8-t)^1/2

then I plugged in the numbers

(t+5)(2(8-t))^1/2 = integral (8-t)^1/2

I don't really know what to do could someone help me out? /:
• Feb 3rd 2013, 07:31 PM
Prove It
Re: Using integral by parts
Integration by Parts is not the best method to use here.

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

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

\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.
• Feb 3rd 2013, 08:03 PM
kuppina
Re: Using integral by parts
Quote:

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

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

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

\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.