<s>[INST] Help with integral from Apostol I've been working on this all day and I'm still stumped. To state the problem (ref: Apostol Section 5.11, Question 29): 
Show
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{2}{a(2n+3)}(x^n(ax+b)^{3/2} - nb\int x^{n-1}\sqrt{ax+b}\,\,dx) + c$$
My attempts so far have been as follows (NB: Im avoiding using integrating with partial fractions because Im not up to that stuff in Apostol):


*

*Let $w = \sqrt{ax+b}$ which lead to
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{2}{a^{n+1}}\int w^2 (w^2 - b)^n\, dw
$$
and then use integration by parts with $u = w$ and $dv = (\frac{w^2-b}{a})^{n} \frac{2w}{a}$ to get
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{1}{n+1}x^{n+1}\sqrt{ax+b} - \frac{1}{n+1}\int x^{n+1}\frac{a}{2\sqrt{ax+b}} dx
$$

*Use integration by parts directly with $u = x^n$ and $dv =\sqrt{ax+b}\,dx$ to get
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{2}{3a}(x^n(ax+b)^{3/2} - n\int\,x^{n-1}(ax+b)^{3/2}\,dx
) + c
$$

*Same as (1) but with the integration by parts, let $dv = w^2$ and $u = (\frac{w^2-b}{a})^{n}$ to get
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{2}{3a}(x^n(ax+b)^{3/2} - n\int\,x^{n-1}(ax+b)^{3/2}\,dx
) + c
$$

*Using integration by parts using $u = x^n\sqrt{ax+b}$ which lead to
$$
\int x^n\sqrt{ax+b}\,\,dx = \frac{x^{n+1}}{n+1}\sqrt{ax+b} - \frac{a}{2(n+1)}\int x^{n+1}(ax+b)^{-1/2}\, dx
$$
which doesn't help because evaluating the integral on the RHS would cancel each side out.

As always, not looking for an answer but some broad hints (so I learn more!)
 [/INST] I would start with integration by parts with the choice $$u = x^n, \quad dv = (ax+b)^{1/2} \, dx,$$ which gives $$du = nx^{n-1}, \quad v = \frac{2}{3a}(ax+b)^{3/2}.$$  Thus if $I$ is the given integral, then $$\begin{align*} I &= \int u \, dv = uv - \int v \, du \\ &= \frac{2}{3a}x^n (ax+b)^{3/2} - \frac{2n}{3a} \int x^{n-1} (ax+b)^{3/2} \, dx \\ &= \frac{2}{3a} x^n (ax+b)^{3/2} - \frac{2n}{3a} \int ax^n (ax+b)^{1/2}  + bx^{n-1} (ax+b)^{1/2} \, dx \\ &= \frac{2}{3a} x^n(ax+b)^{3/2} - \frac{2n}{3} I - \frac{2nb}{3a} \int x^{n-1} (ax+b)^{1/2} \, dx . \end{align*}$$  Therefore, $$\Bigl(\frac{2n}{3}+1\Bigr)I = \frac{2}{3a}x^n(ax+b)^{3/2} - \frac{2nb}{3a} \int x^{n-1} (ax+b)^{1/2} \, dx.$$  Dividing both sides by the constant gives the desired identity.
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