Integrating Misc Integral with \int \frac{x}{\sqrt{3-x^4}}dx

[nameVoid]

$$\int \frac{x}{\sqrt{3-x^4}}dx$$
$$u=x^2$$
$$du=2xdx$$
$$\frac{1}{2}\int\frac{du}{\sqrt{3-u^2}}$$
$$u=\sqrt{3}sinT$$
$$du=\sqrt{3}cosTdT$$
$$\frac{1}{2}\int \frac{\sqrt{3}cosT}{\sqrt{3}cosT}dT<br />$$
$$\frac{x}{2}+C$$

 

[Dick]

Not x/2+C. T/2+C.

 

[synapsis]

You need not do the second u-substitution. Use an integral table to solve it once you have done the first u-substitution. You can google 'integral table' to find one if you don't have one in your book.