Evaluate the integral $$\int \sqrt{x} \ln(1+x)dx $$
so we should start with the substitution: $t=\sqrt{x}$
$$ \int t\ln(1+t)dt2t = 2\int t^2\ln(1+t)dt $$
From here, it seems reasonable to integrate by parts;
$$\frac{t^3}{3}\ln(1+t) - \frac{1}{3} \int \frac{t^3}{1+t} dt$$
So basically we left with evaluating $\int \frac{t^3}{1+t}dt$
And I guess the trick here is to do the following algebraic trick:
$$\int \frac{t^3}{1+t}dt = \int \frac{t^3 + t^2 - t^2}{1+t} dt = \int \frac{t^2(t+1) - t^2}{1+t} dt = \int t^2 - \int \frac{t^2}{1+t}$$
And that's leaves us with a similar problem (And I could go fourth of course)
So although it wasn't that tedious.. Is there a quicker way?
Hint:
Using the antiderivative
$$\int\sqrt{x}\,\mathrm{d}x=\frac23x^{3/2}+\color{grey}{constant}$$
and the derivative
$$\frac{d}{dx}\ln{(1+x)}=\frac{1}{1+x},$$
we can integrate by parts to find,
$$\int\sqrt{x}\,\ln{\left(1+x\right)}\,\mathrm{d}x=\frac23x^{3/2}\,\ln{\left(1+x\right)}-\int\frac{\frac23x^{3/2}}{1+x}\,\mathrm{d}x.$$
Integrating by parts first like this is probably the quicker route.