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Verify the given integral formula

Verify the formula using any method.

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

for n \neq -1.

Proof. To accomplish this, we start with a slightly different integral and (eventually) recover a formula for the integral we want,

    \[ \int \frac{dx}{x^{n-1} \sqrt{ax+b}}, \]

and integrate by parts, letting

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

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*}

Now, we solve for the integral \int \frac{dx}{x^n \sqrt{ax+b}} \, dx,

    \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*}


  1. Mihajlo says:

    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).

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