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Integration by parts.

paulmdrdo

Active member
May 13, 2013
386
how would i go about solving these problems?

\begin{align*}\displaystyle \int\frac{xe^x}{(1+x)^2}dx\end{align*}

\begin{align*}\int\frac{(1-x)dx}{\sqrt{1-x^2}}\end{align*}

this is my solution to prob 2

\begin{align*}\displaystyle\int\frac{(1-x)dx}{\sqrt{1-x^2}}\,=\,\int\frac{dx}{\sqrt{1-x^2}}-\int\frac{x}{\sqrt{1-x^2}}dx\end{align*}

then,
\begin{align*}\displaystyle \sin^{-1}x+C_1-\int\frac{x}{\sqrt{1-x^2}}dx\\\\u\,=\,1-x^2\\du\,=\,-2xdx\\dx\,=\,\frac{du}{-2x}\\\\\int\frac{x}{\sqrt{u}(-2x)}du\\\\-\frac{1}{2}\int\frac{du}{\sqrt{u}}=\,-\frac{1}{2}\frac{(u^{-\frac{1}{2}})}{\frac{1}{2}}\\\\ \sin^{-1}x+(1-x^2)^{\frac{1}{2}}+C\end{align*}

is my answer correct in prob 2?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
For problem 2, you have done well to split the integral:

\(\displaystyle \int\frac{1-x}{\sqrt{1-x^2}}\,dx=\int\frac{1}{\sqrt{1-x^2}}\,dx-\int\frac{x}{\sqrt{1-x^2}}\,dx\)

You have done well also to recognize that:

\(\displaystyle \int\frac{1}{\sqrt{1-x^2}}\,dx=\sin^{-1}(x)+C\)

and your substitution for the second integral is a good one:

\(\displaystyle u=1-x^2\,\therefore\,du=-2x\,dx\)

and so we now have:

\(\displaystyle -\int\frac{x}{\sqrt{1-x^2}}\,dx=\frac{1}{2}\int u^{-\frac{1}{2}}\,du\)

Your methodology for doing substitutions is a bit different than mine, but you have the correct result.

For the first one, I suggest adding to the numerator of the integrand:

\(\displaystyle 0=e^x-e^x\)

and after the correct factorization, see if you find the product rule of differentiation can be applied to the integrand. :D

edit: what happens if in trying integration by parts, you decide to let:

\(\displaystyle u=\frac{e^x}{x+1}\)

What is $du$?
 

paulmdrdo

Active member
May 13, 2013
386
this is what i tried

\begin{align*} \displaystyle \int\frac{xe^x}{(1+x)^2}dx\,=\,\int\frac{xe^x+e^x-e^x}{(1+x)^2}dx\\\\ \int\frac{e^x(x+1-1)}{(1+x)^2}dx\end{align*}

i couldn't continue. :confused:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Try:

\(\displaystyle \frac{(x+1)e^x-e^x}{(1+x)^2}\)

Doesn't this look like it could be the result of differentiating a quotient?
 

paulmdrdo

Active member
May 13, 2013
386
$\displaystyle \frac{(x+1)e^x-e^x}{(1+x)^2}$ - is this equivalent to the original integrand? uhhmm sorry still don't get it.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$\displaystyle \frac{(x+1)e^x-e^x}{(1+x)^2}$ - is this equivalent to the original integrand? uhhmm sorry still don't get it.
Yes it is...we have simply added \(\displaystyle 0=e^x-e^x\) to the numerator.

\(\displaystyle \frac{xe^x}{(1+x)^2}=\frac{xe^x+e^x-e^x}{(1+x)^2}=\frac{\left(xe^x+e^x \right)-e^x}{(1+x)^2}=\frac{(x+1)e^x-e^x}{(1+x)^2}\)
 

paulmdrdo

Active member
May 13, 2013
386
$ \displaystyle \int \frac{(xe^x+e^x)-e^x}{(1+x)^2}dx $

$ \displaystyle \int \frac{(x+1)e^x-e^x}{(1+x)^2}dx \,=\,\int \frac{e^x}{(1+x)}dx-\int \frac{e^x}{(1+x)^2}dx $

what's next?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
This is what I was trying to get you to see:

\(\displaystyle \frac{(x+1)e^x-e^x}{(1+x)^2}=\frac{(x+1)\frac{d}{dx}\left(e^x \right)-e^x\frac{d}{dx}(1+x)}{(1+x)^2}\)

Now, isn't this the result of the differentiation of a quotient?
 

paulmdrdo

Active member
May 13, 2013
386
if that is the case the answer would be $ \displaystyle \frac{e^x}{(1+x)} $?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Minus the constant of integration, yes. Try differentiating this and see what you get.
 

paulmdrdo

Active member
May 13, 2013
386
ahmm.. that solution is kind of subtle. is there another way of solving that?

$\displaystyle D_x(\frac{e^x}{(1+x)})\,=\, \frac{(x+1)d_xe^x-e^x\,d_x(1+x)}{(1+x)^2}$

$\displaystyle \frac{(x+1)e^x-e^x}{(1+x)^2}$ this is equal to the original integrand.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
ahmm.. that solution is kind of subtle. is there another way of solving that?
Yes, again we will use:

\(\displaystyle \int\frac{xe^x}{(1+x)^2}\,dx=\int\frac{(x+1)e^x-e^x}{(1+x)^2}\,dx=\int\frac{e^x}{(1+x)}\,dx-\int\frac{e^x}{(1+x)^2}\,dx\)

Now, for the first integral, using integration by parts, we find:

\(\displaystyle u=\frac{1}{1+x}\,\therefore\,du=-\frac{1}{(1+x)^2}\)

\(\displaystyle dv=e^x\,dx\,\therefore\,v=e^x\)

and so we have:

\(\displaystyle \int\frac{xe^x}{(1+x)^2}\,dx= \frac{e^x}{1+x}+\int\frac{e^x}{(1+x)^2}\,dx-\int\frac{e^x}{(1+x)^2}\,dx\)

Hence:

\(\displaystyle \int\frac{xe^x}{(1+x)^2}\,dx=\frac{e^x}{1+x}+C\)
 

paulmdrdo

Active member
May 13, 2013
386
mark on the first integral why there's an x in the numerator? i suppose it's e^x only.
$\displaystyle \int\frac{xe^x}{(1+x)^2}\,dx=\int\frac{(x+1)e^x-e^x}{(1+x)^2}\,dx=\int\frac{xe^x}{(1+x)}\,dx-\int\frac{e^x}{(1+x)^2}\,dx$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, that was a typo, which I have fixed.