How can I simplify this integral with trigonometric functions involved?

[ls_boy]

I'm trying to understand this integral.

$\int \frac {x^3 dx}{x^4 + 1}dx\\$

One possible answer is this formula.
$\int \frac {dx}{x}dx = \ln |x|+ C\\$

I've seen a couple of other formulas where if $x^2 + a^2$ is a part of the integrand
than trigonometric functions become part of the end formula. There are too many
formulas to choose from.

 

[dextercioby]

Can you make a substitution ? Write \ln in LaTex to get a nice appearance of the function.

 

[saminator910]

The key here is finding the proper substitution

$u = x^{4}+1$

$du = 4x^{3}dx$

sub in

$\displaystyle\frac{1}{4}\int \frac{du}{u}$

$\displaystyle\frac{\ln u}{4}+c$

re-sub

$\displaystyle\frac{\ln (x^{4}+1)}{4}+c$

 

[ls_boy]

Thanks!

 

[D H]

The key here is finding the proper substitution

$u = x^{4}+1$

$du = 4x^{3}dx$

There are other ways to solve this problem. Yours happens to be the easiest.

For example, you could start with the u-substitution $u=x^2$. With this, the integral becomes $\int \frac {x^3}{x^4+1}\,dx = \frac 1 2 \int \frac{u}{u^2+1}\,du$. From here, you could recognize that this is of the form $\int \frac {dx}{x}$ and get the same result as you did (but with an extra step).

Let's say one recognized this as instead needing a trig substitution. The $u^2+1$ denominator suggests using $u = \tan \theta$. With this, the integral becomes $\int \frac{u}{u^2+1}\,du = \int \frac{\tan \theta}{\tan^2\theta+1}\sec^2\theta\,d\theta = \int \tan\theta \,d\theta$. Some people will immediately recognize that as being $-\log(\cos\theta)$. For those that don't know this off the top of their heads, they'd have to finally recognize that $\int \tan\theta \,d\theta = \int \frac{\sin \theta}{\cos\theta}\,d\theta$ is of the form $\int \frac {dx}{x}$.

Now you have to substitute back so the integral is expressed in terms of $u$ rather than $\theta$, and then in terms of $x$ rather than $u$. Eventually you'll get $\log(x^4+1)/4$ as the result, the same result as the easy route. But getting there by this trig route is a lot more torturous than recognizing the easy form right off the bat.

 

[micromass]

A trick that you should certainly remember is

$$\int \frac{f^\prime(x)}{f(x)}dx = \log|f(x)| + C$$

This is easily proven by differentiating the RHS.

It will make your life simpler in many occasions (including now), and sometimes it is the only way to solve the problem!

Of course, this comes down to what the previous posters have said, but it's still a good trick to know.