1. ## definite integral

find the indefinite integral of x/(1+x^4)

Let u = x^2; then du = 2x dx, so

∫ x dx /(1+x^4) = (1/2) ∫ 2x dx / (1+x^4) = (1/2) ∫ du / (1 + u^2)......blanked out

ans: arctan (x^2)/2....huh??

2. Originally Posted by Taurus3
find the indefinite integral of x/(1+x^4)

Let u = x^2; then du = 2x dx, so

∫ x dx /(1+x^4) = (1/2) ∫ 2x dx / (1+x^4) = (1/2) ∫ du / (1 + u^2)......blanked out

ans: arctan (x^2)/2....huh??
You are good so far now make the trig sub

$u=\tan(\theta) \implies du =\sec^2{\theta}d\theta$ and remember that

$\tan^2(\theta)+1=\sec^2{\theta}$

3. hold, on. I don't get what you mean.

4. Originally Posted by Taurus3
hold, on. I don't get what you mean.
just plug in what I suggested to get

$\displaystyle \frac{1}{2} \int \frac{du}{1+u^2}=\frac{1}{2} \int \frac{\sec^2(\theta)}{1+\tan^2(\theta)}d\theta=\fr ac{1}{2} \int d\theta=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}(u)=...$

5. oh dumb....thanks haha.