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Verify the given integral formula

Verify the formula using any method.

    \[ \int x^n \sqrt{ax+b} \, dx = \frac{2}{a(2n+3)} \left( x^n (ax+b)^{\frac{3}{2}} - nb \int x^{n-1} \sqrt{ax+b} \, dx \right) + C \]

for n \neq -\frac{3}{2}.


Proof. First, we integrate by parts with

    \begin{align*}  u &= x^n & du &= nx^{n-1} \\  dv &= \sqrt{ax+b}  \, dx & v &= \frac{2}{3a}(ax+b)^{\frac{3}{2}}.  \end{align*}

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*}

Now, we bring the \frac{2n}{3} \int x^n \sqrt{ax+b} \, dx 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*}

This was the requested integral formula. \qquad \blacksquare

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