1. Intergration: Substitution - Question

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

I first started by setting $\displaystyle u = 1+x^4$ ,but then $\displaystyle du = 4x^3dx$ , doesn't work out with the $\displaystyle xdx$ that is left.

Any ideas?

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$\displaystyle \int \frac{x}{1+x^2} \ dx$

I first started by setting $\displaystyle u = 1+x^4$ ,but then $\displaystyle du = 4x^3dx$ , doesn't work out with the $\displaystyle xdx$ that is left.
Let,
$\displaystyle u=1+x^2$ then, $\displaystyle \frac{du}{dx}=2x$
Therefore, express integral as,
$\displaystyle \frac{1}{2}\int \frac{1}{1+x^2}\cdot (2x) dx$
Substitute,
$\displaystyle \frac{1}{2}\int \frac{1}{u}\frac{du}{dx} dx=\frac{1}{2}\int \frac{1}{u} du=\frac{1}{2}\ln|u|+C$Thus,
$\displaystyle \frac{1}{2}\ln |1+x^2|+C$
but, $\displaystyle 1+x^2>0$ therefore,
$\displaystyle \frac{1}{2}\ln (1+x^2)+C$

3. Oh, sorry, I made a mistake, the problem should be:

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

Is to the 4th power instead of 2nd.

Oh, sorry, I made a mistake, the problem should be:

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

Is to the 4th power instead of 2nd.
You have,
$\displaystyle \int \frac{x}{1+x^4}dx$
Let,
$\displaystyle u=x^2$ then, $\displaystyle \frac{du}{dx}=2x$
Express integrand as,
$\displaystyle \frac{1}{2}\int \frac{1}{1+(x^2)^2}\cdot (2x) dx$ and substitute,
$\displaystyle \frac{1}{2}\int \frac{1}{1+u^2}\frac{du}{dx}dx=\frac{1}{2}\int \frac{du}{1+u^2}=\frac{1}{2}\tan^{-1}(u)+C$
Substitute back,
$\displaystyle \frac{1}{2}\tan^{-1}(x^2)+C$