# Integral of $\int^{\infty}_0\frac{x}{x^4+1}\,dx$

I want to evaluate this integral and trying to figure how to do that. $$\int^{\infty}_0\frac{x}{x^4+1}dx$$
What I did is:
Im abandons the limits for now and do the following step: $$\int^{\infty}_0\frac{x}{x^4+1}dx = \frac{1}{2}\int\frac{tdt}{t^2+1}$$ I can use this integral? $\rightarrow$ $\int \frac{x}{x^2+a}dx = \frac{ln|x^2+a|}{2}+C$? or I can do it another way?
thanks!

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You made an error using substitution. – Vincent Pfenninger Jun 8 '13 at 15:34

Hint: Let $u=x^2$ and write your integral in the form

$$\int^{\infty}_0\frac{x}{x^4+1}dx = \int_0^{\infty} \frac{1}{(x^2)^2+1}x\,dx.$$

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what Im doing with the upper $x$ I dont understand it. if I choose $x^2$ to be $u$ so I have $$du = 2xdx$$ and I need to find an expression to the upper $x$ and the expression for $x$ is $\sqrt{u}$ no? – Ofir Attia Jun 8 '13 at 15:42
Never mind, I found my mistake. thanks. – Ofir Attia Jun 8 '13 at 15:51

$$\int\frac{1}{1+(x^2)^2} {\huge(} x\,dx {\huge)} = \frac12\int\frac{1}{1+(x^2)^2} {\huge(} 2x\,dx {\huge)}$$

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I think what you want is

$$\int_0^{\infty} dx \frac{x}{x^4+1} = \frac12 \int_0^{\infty} \frac{dx}{x^2+1} = \frac{\pi}{4}$$

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Putting $x^2=\tan\theta,$

$2x\,dx=\sec^2\theta \,d\theta$ and $x^4+1=(\tan\theta)^2+1=\sec^2\theta$

$$\int^{\infty}_0\frac{x}{x^4+1} \, dx=\int_0^{\frac\pi2}\frac{\sec^2\theta \, d\theta}{2\sec^2\theta }=\frac12\int_0^{\frac\pi2} \, d\theta$$

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