Question

Integrate from 1 to 3:
(2x^2+2)/sqrt(x^3+3x)

Answer 1

well I'd let

\[u = x^3 + 3x\]

so the boundary values become

4 to 36

then

\[\frac{du}{dx} = 3x^2 + 3\]

to get \[2x^2 + 2\]

you need to let

\[\frac{2}{3}du = (2x^2 + 2) dx\]

so your problem is now

\[\int\limits_{4}^{36} u^{-\frac{1}{2}} \times \frac{2}{3} du\]

which can be written as \[\frac{2}{3} \int\limits_{4}^{36} u^{-\frac{1}{2}} du\]

now integrate with respect to u

and evaluate