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|??
Hi
You can either substitute $\displaystyle t=\exp x$ or remember that $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}(\arcsin u(x))=\frac{u'(x)}{\sqrt{1-u^2(x)}}$
You may rewrite it as $\displaystyle \frac{3-x}{x^2+4}=\frac{3}{x^2+4}-\frac{x}{x^2+4}2) (3-x) / (x^2+4)
=\frac{3}{4}\frac{1}{1+\left(\frac{x}{2}\right)^2}-\frac{1}{2}\frac{2x}{x^2+4}$
No, that's wrong. (except if $\displaystyle u(x)=x$ )Also I got one intergral down to 1/u du, and is 1/u the same as ln|u|??
1) $\displaystyle \int\frac{3e^x}{\sqrt{1-e^{2x}}}\,dx$
You can rewrite the integral as follows:
$\displaystyle \int\frac{3e^x}{\sqrt{1-(e^x)^2}}\,dx$
Now our substitution will become evident. Let $\displaystyle u=e^x$. $\displaystyle du=e^x dx\rightarrow dx=\frac{du}{e^x}$.
Substitution into the integral, we have:
$\displaystyle \int\frac{3e^x}{\sqrt{1-u^2}}\frac{du}{e^x}=3\int\frac{du}{\sqrt{1-u^2}}$
The integral of this is $\displaystyle sin^{-1}u+C$. Substitute the u value back in, and you get:
$\displaystyle \int\frac{3e^x}{\sqrt{1-e^{2x}}}\,dx=3sin^{-1}{\left(e^x\right)}+C$
2) $\displaystyle \int\frac{3-x}{x^2+4}\,dx$
I would recommend rewriting the integrand and then break it up into 2 integrals.
$\displaystyle \int\frac{3-x}{x^2+4}\,dx=\int\frac{3}{x^2+4}\,dx-\int\frac{x}{x^2+4}\,dx$.
The first integral will take on the form of $\displaystyle tan^{-1}x$, in particular, it will be $\displaystyle \frac{1}{2}tan^{-1}{\left(\frac{x}{2}\right)}+C$.
The second integral requires a substitution: $\displaystyle u=x^2+4$. $\displaystyle du=2x\rightarrow dx=\frac{du}{2x}$. Substituting into the second integral, we have:
$\displaystyle \int\frac{x}{u}\frac{du}{2x}=\frac{1}{2}\int\frac{ du}{u}$.
This integral wll be $\displaystyle ln{\left|u\right|}+C$. Substitute u back in and you get $\displaystyle ln{\left|x^2+4\right|}+C$.
Thus:
$\displaystyle \int\frac{3}{x^2+4}\,dx-\int\frac{x}{x^2+4}\,dx=\frac{3}{2}tan^{-1}{\left(\frac{x}{2}\right)}-ln{\left|x^2+4\right|}+C$
Hope this answers your question!!!
Yes, splitting it into two integrals is a good idea :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?
$\displaystyle \int \frac{3-x}{x^2+4}\,\mathrm{d}x=\frac{3}{2}\int \frac{\frac{1}{2}}{1+\left(\frac{x}{2}\right)^2}\, \mathrm{d}x-\frac{1}{2}\int\frac{2x}{x^2+4}\,\mathrm{d}x$
For the first one, you can try $\displaystyle X=\frac{x}{2}$ and for the second one, $\displaystyle u=x^2+4$. Otherwise, I wrote them so that you can integrate directly if you "see" the derivatives which are involved.
$\displaystyle \int \frac{3-x}{x^2+4}\,\mathrm{d}x$
$\displaystyle = \int \frac{3}{x^2+4}\,\mathrm{d}x - \int \frac{x} {x^2+4}\,\mathrm{d}x$
$\displaystyle = 3 \int \frac{1}{x^2+4}\,\mathrm{d}x - \int \frac{x} {x^2+4}\,\mathrm{d}x$
First integration is just a standard integration ($\displaystyle \mathrm{arctan}$) and the second one, you could use substitution with $\displaystyle u=x^2 + 4$.