# Math Help - 2 more intergrals

1. ## 2 more intergrals

Last two intergrals and I'm done for today's homework.

1) 3e^x / sqrt(1-e^(2x))

2) (3-x) / (x^2+4)

Also I got one intergral down to 1/u du, and is 1/u the same as ln|u|??

2. Hi
Originally Posted by Hibijibi
1) 3e^x / sqrt(1-e^(2x))
You can either substitute $t=\exp x$ or remember that $\frac{\mathrm{d}}{\mathrm{d}x}(\arcsin u(x))=\frac{u'(x)}{\sqrt{1-u^2(x)}}$
2) (3-x) / (x^2+4)
You may rewrite it as $\frac{3-x}{x^2+4}=\frac{3}{x^2+4}-\frac{x}{x^2+4}
=\frac{3}{4}\frac{1}{1+\left(\frac{x}{2}\right)^2}-\frac{1}{2}\frac{2x}{x^2+4}$

Also I got one intergral down to 1/u du, and is 1/u the same as ln|u|??
No, that's wrong. (except if $u(x)=x$ )

3. I solved the first one, but the second still confused me a bit on how to approach it. Should I split into two separate integrals? And what should u be for those?

4. Originally Posted by Hibijibi
Last two intergrals and I'm done for today's homework.

1) 3e^x / sqrt(1-e^(2x))

2) (3-x) / (x^2+4)

Also I got one intergral down to 1/u du, and is 1/u the same as ln|u|??
1) $\int\frac{3e^x}{\sqrt{1-e^{2x}}}\,dx$

You can rewrite the integral as follows:

$\int\frac{3e^x}{\sqrt{1-(e^x)^2}}\,dx$

Now our substitution will become evident. Let $u=e^x$. $du=e^x dx\rightarrow dx=\frac{du}{e^x}$.

Substitution into the integral, we have:

$\int\frac{3e^x}{\sqrt{1-u^2}}\frac{du}{e^x}=3\int\frac{du}{\sqrt{1-u^2}}$

The integral of this is $sin^{-1}u+C$. Substitute the u value back in, and you get:

$\int\frac{3e^x}{\sqrt{1-e^{2x}}}\,dx=3sin^{-1}{\left(e^x\right)}+C$

2) $\int\frac{3-x}{x^2+4}\,dx$

I would recommend rewriting the integrand and then break it up into 2 integrals.

$\int\frac{3-x}{x^2+4}\,dx=\int\frac{3}{x^2+4}\,dx-\int\frac{x}{x^2+4}\,dx$.

The first integral will take on the form of $tan^{-1}x$, in particular, it will be $\frac{1}{2}tan^{-1}{\left(\frac{x}{2}\right)}+C$.

The second integral requires a substitution: $u=x^2+4$. $du=2x\rightarrow dx=\frac{du}{2x}$. Substituting into the second integral, we have:

$\int\frac{x}{u}\frac{du}{2x}=\frac{1}{2}\int\frac{ du}{u}$.

This integral wll be $ln{\left|u\right|}+C$. Substitute u back in and you get $ln{\left|x^2+4\right|}+C$.

Thus:

$\int\frac{3}{x^2+4}\,dx-\int\frac{x}{x^2+4}\,dx=\frac{3}{2}tan^{-1}{\left(\frac{x}{2}\right)}-ln{\left|x^2+4\right|}+C$

5. I solved the first one, but the second still confused me a bit on how to approach it. Should I split into two separate integrals? And what should u be for those?
Yes, splitting it into two integrals is a good idea :

$\int \frac{3-x}{x^2+4}\,\mathrm{d}x=\frac{3}{2}\int \frac{\frac{1}{2}}{1+\left(\frac{x}{2}\right)^2}\, \mathrm{d}x-\frac{1}{2}\int\frac{2x}{x^2+4}\,\mathrm{d}x$

For the first one, you can try $X=\frac{x}{2}$ and for the second one, $u=x^2+4$. Otherwise, I wrote them so that you can integrate directly if you "see" the derivatives which are involved.

6. Originally Posted by Hibijibi
I solved the first one, but the second still confused me a bit on how to approach it. Should I split into two separate integrals? And what should u be for those?
$\int \frac{3-x}{x^2+4}\,\mathrm{d}x$

$= \int \frac{3}{x^2+4}\,\mathrm{d}x - \int \frac{x} {x^2+4}\,\mathrm{d}x$

$= 3 \int \frac{1}{x^2+4}\,\mathrm{d}x - \int \frac{x} {x^2+4}\,\mathrm{d}x$

First integration is just a standard integration ( $\mathrm{arctan}$) and the second one, you could use substitution with $u=x^2 + 4$.