Verify the formula using any method.
![\[ \int \frac{x^m}{\sqrt{a+bx}} \, dx = \frac{2}{(2m+1)b} \left( x^m \sqrt{a+bx} - ma \int \frac{x^{m-1}}{\sqrt{a+bx}} \, dx \right) + C \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-579d9c507d71f9e215ca2565f6a96124_l3.png "Rendered by QuickLaTeX.com")
for 
.
Proof. First, we use integration by parts, letting
This gives us
![\begin{align*} \int \frac{x^m}{\sqrt{a+bx}} \, dx &= uv - \int v \, du \\[9pt] &= \frac{2x^m}{b} \sqrt{a+bx} - \frac{2m}{b} \int x^{m-1} \sqrt{a+bx} \, dx \\[9pt] &= \frac{2x^m}{b} \sqrt{a+bx} - \frac{2m}{b} \int \frac{x^{m-1} (a+bx)}{\sqrt{a+bx}} \, dx \\[9pt] &= \frac{2x^m}{b} \sqrt{a+bx} - \frac{2ma}{b} \int \frac{x^{m-1}}{\sqrt{a+bx}} \, dx - 2m \int \frac{x^m}{\sqrt{a+bx}} \, dx. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-78040022eb8ee699e649f27178f43584_l3.png "Rendered by QuickLaTeX.com")
Now, we bring the 
back to the left since this is the same integral that we started with. This gives us,
![\begin{align*} && (2m+1) \int \frac{x^m}{\sqrt{a+bx}} \, dx &= \frac{2x^m}{b} \sqrt{a+bx} - \frac{2ma}{b} \int \frac{x^{m-1}}{\sqrt{a+bx}} \, dx \\[9pt] \implies && \int \frac{x^m}{\sqrt{a+bx}} \, dx &= \frac{2}{(2m+1)b} \left(x^m \sqrt{a+bx} - ma \int \frac{x^{m-1}}{\sqrt{a+bx}} \ dx\right) + C . \qquad \blacksquare \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-55515b954d596e912c6ff2485dabc80f_l3.png "Rendered by QuickLaTeX.com")
Verify the formula using any method.
for .
Proof. First, we use integration by parts, letting
This gives us
Now, we bring the back to the left since this is the same integral that we started with. This gives us,