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Verify the given integral formula

Verify the formula using any method.

    \[ \int \frac{x^m}{\sqrt{a+bx}} \, dx = \frac{2}{(2m+1)b} \left( x^m \sqrt{a+bx} - ma \int \frac{x^{m-1}}{\sqrt{a+bx}} \, dx \right) + C \]

for n \neq -\frac{1}{2}.


Proof. First, we use integration by parts, letting

    \begin{align*}  u &= x^m & du &= m x^{m-1} \\ dv &= \frac{1}{\sqrt{a+bx}} \, dx & v &= \frac{2}{b} \sqrt{a+bx}. \end{align*}

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*}

Now, we bring the 2m \int \frac{x^m}{\sqrt{a+ bx}} \, dx 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*}

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