MHB Integral: $\int \frac{x}{9+x^4}dx$

[karush]

$\int\frac{x}{9+x^4}dx$
$$u=x^2\ du=2x\ dx\ \ x=\sqrt{x}$$
I assume this going to have a trig answer but I didn't know how to deal with the $$dx$$

 

[mathmari]

$\int\frac{x}{9+x^4}dx$
$$u=x^2\ du=2x\ dx\ \ x=\sqrt{x}$$
I assume this going to have a trig answer but I didn't know how to deal with the $$dx$$

$$u=x^2 \Rightarrow du=2xdx$$

$$\int\frac{x}{9+x^4}dx=\frac{1}{2}\int \frac{1}{9+u^2}du=\frac{1}{2 \cdot 9} \int \frac{1}{1+\left (\frac{u}{3}\right)^2 }$$

$$\frac{u}{3}=\tan w \Rightarrow \frac{1}{3}du=\frac{1}{\cos^2 w} dw$$

$$\frac{1}{1+\tan^2 w}=\cos^2 w$$

$$\frac{1}{2 \cdot 9} \int \frac{1}{1+\left (\frac{u}{3}\right)^2 }du=\frac{1}{6} \int dw=\frac{w}{6}+c=\frac{arc \tan \frac{u}{3}}{6}+c=\frac{arc \tan \frac{x^2}{3}}{6}+c$$

Therefore, $$\int\frac{x}{9+x^4}dx=\frac{arc \tan \frac{x^2}{3}}{6}+c$$

 

[karush]

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[Jameson]

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[karush]

OK thank you