How Can I Solve These Integration Problems Using Integration by Parts?

[paulmdrdo1]

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?

 

[MarkFL]

For problem 2, you have done well to split the integral:

$$\int\frac{1-x}{\sqrt{1-x^2}}\,dx=\int\frac{1}{\sqrt{1-x^2}}\,dx-\int\frac{x}{\sqrt{1-x^2}}\,dx$$

You have done well also to recognize that:

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

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

$$u=1-x^2\,\therefore\,du=-2x\,dx$$

and so we now have:

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

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:

$$0=e^x-e^x$$

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:

$$u=\frac{e^x}{x+1}$$

What is $du$?

 

[paulmdrdo1]

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.

 

[MarkFL]

Try:

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

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

 

[paulmdrdo1]

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

 

[MarkFL]

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

Yes it is...we have simply added $$0=e^x-e^x$$ to the numerator.

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

 

[paulmdrdo1]

$ \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]

This is what I was trying to get you to see:

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

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

 

[paulmdrdo1]

if that is the case the answer would be $ \displaystyle \frac{e^x}{(1+x)} $?

 

[MarkFL]

Minus the constant of integration, yes. Try differentiating this and see what you get.

 

[paulmdrdo1]

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]

ahmm.. that solution is kind of subtle. is there another way of solving that?

Yes, again we will use:

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

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

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

$$dv=e^x\,dx\,\therefore\,v=e^x$$

and so we have:

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

Hence:

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

 

[paulmdrdo1]

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]

Yes, that was a typo, which I have fixed.