Q:

Integration by using a suitable substitution

I'm trying to integrate $\int_2^t x^3/\sqrt{x^4 - 2}$. I need to use a suitable substitution method; I tried letting $u=(x^4 - 2)$. I got the answer of $-1/4(\sqrt{t^4 - 2} - \sqrt{14})$; the actual answer apparently is the same but with 1/2 instead of -1/4.
Can someone help?

A:

Use the general formula for a differentiable function $\;f(x)\;$, which you can easily prove using the Chain Rule :
$$\int \frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}+C$$
And then, since $\;(x^4-2)'=4x^3\;$ , we get
$$\int\limits_2^t\frac{x^3}{\sqrt{x^4-2}}dx=\frac14\int\limits_2^t\frac{4x^3}{\sqrt{x^4-2}}dx=\left.\frac12\sqrt{x^4-2}\right|_2^t=\frac12\left(\sqrt{t^4-2}-\sqrt{14}\right)$$

