Question

integral of sqrt(x)/(1+x^3)

Answer 1

For this question I would normally just use integration by parts, but it doesn't seem obvious what I should "keep" and what I should "integrate". Thus I will employ a combination of integration by parts, substitution and chain rule.

First, I would try to do what integration by parts teach us. Integration one part first and then do some other steps.

If I integrate the numerator only of \[\frac{\sqrt{x}}{1+x^3}\]
We will obtain
\[\frac{2}{3}x^{\frac{3}{2}}\]

I look at the x^(3/2) and see that if I squared it, I will get x^3 which appears in the bottom. This makes the integral into something like 1/(1+u^2). It then occurred to me that there is an integration identity for such and integral namely: \[\int\frac{1}{1+u^2}du = \frac{1}{a}tan^{-1}\frac{u}{a}\]

But first let me do a trick chain rule taught us:
\[\int u\:\frac{dv}{dx}\:dx = \int u\:dv\]
So what we do is make one part of it dv/dx, while the other be u. In this case, I find letting dv/dx be sqrt(x) easier, so dv = d(2/3 x^3/2)

\[\int\frac{\sqrt{x}}{1+x^3}dx = \int\frac{1}{1+x^3}d(\frac{2}{3}x^{3/2})\]

Putting the constants aside, and substituting x^3/2 = u for better clarity:
\[\frac{2}{3}\int\frac{1}{1+u^2}du\]
Now using the integration identity where a = 1
\[\frac{2}{3}\int\frac{1}{1+u^2}du=\frac{2}{3}tan^{-1}(u)=\frac{2}{3}tan^{-1}x^{\frac{3}{2}}\]

Where tan-1 is arctan.