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??
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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 $\displaystyle u=\tan(\theta) \implies du =\sec^2{\theta}d\theta$ and remember that $\displaystyle \tan^2(\theta)+1=\sec^2{\theta}$
hold, on. I don't get what you mean.
Originally Posted by Taurus3 hold, on. I don't get what you mean. just plug in what I suggested to get $\displaystyle \displaystyle \frac{1}{2} \int \frac{du}{1+u^2}=\frac{1}{2} \int \frac{\sec^2(\theta)}{1+\tan^2(\theta)}d\theta=\fr ac{1}{2} \int d\theta=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}(u)=...$
oh dumb....thanks haha.
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