Verify the formula using any method.
![\[ \int \frac{dx}{x^n \sqrt{ax+b}} = - \frac{\sqrt{ax+b}}{(n-1)bx^{n-1}} - \frac{(2n-3)a}{(2n-2)b} \int \frac{dx}{x^{n-1} \sqrt{ax+b}} + C \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-5bb191f7ce690b5e6def44918588ea19_l3.png "Rendered by QuickLaTeX.com")
for 
.
Proof. To accomplish this, we start with a slightly different integral and (eventually) recover a formula for the integral we want,
![\[ \int \frac{dx}{x^{n-1} \sqrt{ax+b}}, \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-23eae5b4279ed6e88a54d36417a33ebd_l3.png "Rendered by QuickLaTeX.com")
and integrate by parts, letting
This gives us,
![\begin{align*} \int \frac{dx}{x^{n-1} \sqrt{ax+b}} &= uv - \int v \, du \\[9pt] &= \frac{2 \sqrt{ax+b}}{ax^{n-1}} + \frac{2(n-1)}{a} \int \frac{\sqrt{ax+b}}{x^n} \, dx\\[9pt] &= \frac{2 \sqrt{ax+b}}{ax^{n-1}} + \frac{2(n-1)}{a} \int \frac{(ax+b)}{x^n \sqrt{ax+b}} \, dx \\[9pt] &= \frac{2 \sqrt{ax+b}}{ax^{n-1}} + 2(n-1) \int \frac{dx}{x^{n-1}\sqrt{ax+b}} + \frac{2b(n-1)}{a} \int \frac{dx}{x^n \sqrt{ax+b}}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-74b0752a17c1b58146fe863253db35f1_l3.png "Rendered by QuickLaTeX.com")
Now, we solve for the integral 
,
![\begin{align*} &&\frac{2b(n-1)}{a} \int \frac{dx}{x^n \sqrt{ax+b}} &= \int \frac{dx}{x^{n-1} \sqrt{ax+b}} - \frac{2 \sqrt{ax+b}}{ax^{n-1}} - 2(n-1) \int \frac{dx}{x^{n-1} \sqrt{ax+b}} \\[9pt] \implies && \frac{2b(n-1)}{a} \int \frac{dx}{x^n \sqrt{ax+b}} &= -\frac{2\sqrt{ax+b}}{ax^{n-1}} - (2n-3) \int \frac{dx}{x^{n-1} \sqrt{ax+b}} \\[9pt] \implies && \int \frac{dx}{x^n \sqrt{ax+b}} \, dx &= -\frac{\sqrt{ax+b}}{(n-1)b x^{n-1} } - \frac{(2n-3)a}{(2n-2)b} \int \frac{dx}{x^{n-1} \sqrt{ax+b}}. \qquad \blacksquare \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-b7efd9f2301bd2b0f5fdf0827ba64b99_l3.png "Rendered by QuickLaTeX.com")
Integrands are all continuous, so it may be more straightforward to just use that any indefinite integral of a continuous f is also a primitive of f (page 211).
How did you know that solving for x^(n-1) in the first place would lead to the solution?
I alread know why, sorry for not being more attentive.