Verify the formula using any method.
![\[ \int x^n \sqrt{ax+b} \, dx = \frac{2}{a(2n+3)} \left( x^n (ax+b)^{\frac{3}{2}} - nb \int x^{n-1} \sqrt{ax+b} \, dx \right) + C \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-c2421d5c95aa0fb17b8f6fd0082884da_l3.png "Rendered by QuickLaTeX.com")
for 
.
Proof. First, we integrate by parts with
This gives us,
![\begin{align*} \int x^n \sqrt{ax+b} \, dx &= uv - \int v \ du \\[9pt] &= \frac{2}{3a} x^n (ax+b)^{\frac{3}{2}} - \frac{2n}{3a} \int x^{n-1} (ax+b)^{\frac{3}{2}} \, dx \\[9pt] &= \frac{2}{3a} x^n (ax+b)^{\frac{3}{2}} - \frac{2n}{3a} \int x^{n-1} (ax+b) \sqrt{ax+b} \, dx \\[9pt] &= \frac{2}{3a} x^n (ax+b)^{\frac{3}{2}} - \frac{2n}{3a} \int \left( ax^n \sqrt{ax+b} + bx^{n-1} \sqrt{ax+b} \right) \, dx \\[9pt] &= \frac{2}{3a} x^n (ax+b)^{\frac{3}{2}} - \frac{2n}{3} \int x^n \sqrt{ax+b} - \frac{2nb}{3a} \int x^{n-1} \sqrt{ax+b} \, dx. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-610157b887ff93b94cb72d91bde79965_l3.png "Rendered by QuickLaTeX.com")
Now, we bring the 
back over to the left (since this integral is the same one we started with) and have
![\begin{align*} && \left(1 + \frac{2n}{3} \right) \int x^n \sqrt{ax+b} \, dx &= \frac{2}{3a} x^n (ax+b)^{\frac{3}{2}} - \frac{2nb}{3a} \int x^{n-1} \sqrt{ax+b} \, dx \\[9pt] \implies && \frac{2n+3}{3} \int x^n \sqrt{ax+b} \, dx &= \frac{2}{3a} x^n (ax+b)^{\frac{3}{2}} - \frac{2nb}{3a} \int x^{n-1} \sqrt{ax+b} \, dx \\[9pt] \implies && \int x^n \sqrt{ax+b} \, dx &= \frac{2x^n (ax+b)^{\frac{3}{2}}}{a(2n+3)} - \frac{2nb}{a(2n+3)} \int x^{n-1} \sqrt{ax+b} \, dx \\[9pt] \implies && \int x^n \sqrt{ax+b} \, dx &= \frac{2}{a(2n+3)} \left( x^n (ax+b)^{\frac{3}{2}} - nb \int x^{n-1} \sqrt{ax+b} \, dx \right). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-e0e2f735c0a6fa6d48fb9106fc559b4f_l3.png "Rendered by QuickLaTeX.com")
This was the requested integral formula
Indefinite integral of a continuous f is also primitive of f (page 211). That can be used in these exercises to validate the equality, maybe in a more straightforward way.