Calculating Integral of \int \frac{dx}{x(x^4+1)} using Substitution

[nameVoid]

<br /> \int \frac{dx}{x(x^4+1)}<br />
<br /> u=x^2<br />
<br /> \sqrt{u}=x<br />
<br /> dx=\frac{1}{2\sqrt{u}}<br />
<br /> \frac{1}{2}\int \frac{du}{u^2+1}<br />
<br /> \frac{1}{2}arctanx^2+C<br />

 

[Mark44]

<br /> \int \frac{dx}{x(x^4+1)}<br />
<br /> u=x^2<br />
<br /> \sqrt{u}=x<br />
<br /> dx=\frac{1}{2\sqrt{u}}<br />
<br /> \frac{1}{2}\int \frac{du}{u^2+1}<br />

The integral above isn't right. You forgot to replace the x factor in the denominator.

<br /> \frac{1}{2}arctanx^2+C<br />

 

[nameVoid]

<br /> \frac{1}{2} \int \frac{du}{u(u^2+1)}=\int \frac{A}{u}+\frac{Bu+C}{u^2+1}du<br />
<br /> A=\frac{1}{2}=-B<br />
<br /> ln|x|-\frac{1}{4}ln(x^4+1)+C<br />