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Evaluate the integral of (x2 + x)1/2 / x

by RoRi

Compute the following integral.

    \[ \int \frac{\sqrt{x^2+x}}{x} \, dx. \]

First, we have

    \[ \int \frac{\sqrt{x^2+x}}{x} \, dx = \int \frac{\sqrt{x}\sqrt{x+1}}{x} \, dx = \int \sqrt{\frac{x+1}{x}} \, dx. \]

Then, we make a substitution, letting

    \begin{align*} &&u &= \frac{x+1}{x} \\ \implies && du &= -\frac{1}{x^2} \, dx \\ \implies && dx &= \frac{1}{(1-u)^2}. \end{align*}

This gives us

    \[ \int \frac{\sqrt{x^2+x}}{x} \, dx = -\int \frac{\sqrt{u}}{(1-u)^2} \, du. \]

Next, we make another substitution, letting t = \sqrt{u} and dt = \frac{1}{2\sqrt{u}} \, du. This gives us,

    \begin{align*} -\int \frac{\sqrt{u}}{(1-u)^2} \, du &= -2 \int \frac{t^2}{(1-t^2)^2} \, dt \\[9pt] &= -2 \int \frac{t^2}{(t^2-1)^2} \, dt \\[9pt] &= -2 \int \frac{t^2}{(t+1)^2 (t-1)^2} \, dt. \end{align*}

We can then use partial fractions on the integrand,

    \[ \frac{t^2}{(t+1)^2 (t-1)^2} = \frac{A}{t+1} + \frac{B}{(t+1)^2} + \frac{C}{t-1} + \frac{D}{(t-1)^2}. \]

This gives us the equation

    \[ A(t+1)(t-1)^2 + B(t-1)^2 + C(t+1)(t-1) + D(t+1)^2 = t^2. \]

Solving this for A, B, C, and D we obtain

    \[ A = -\frac{1}{4}, \quad B = C = D = \frac{1}{4}. \]

Therefore,

    \begin{align*} -2 \int \frac{t^2}{(t^2-1)^2} \, dt &= -2 \int \left( \frac{-1}{4(t+1)} + \frac{1}{4(t+1)^2} + \frac{1}{4(t-1)} + \frac{1}{4(t-1)^2} \right) \, dt \\[9pt] &= -\frac{1}{2} \left( \int \frac{1}{(t-1)^2} \, dt + \int \frac{1}{t-1} \, dt + \int \frac{1}{(t+1)^2} \, dt - \int \frac{1}{t+1} \, dt \right) \\[9pt] &= \frac{1}{2(t-1)} - \frac{1}{2} \log |t-1| + \frac{1}{2(t+1)} + \frac{1}{2} \log|t+1| + C \\[9pt] &= \frac{1}{2} \left( \frac{1}{t-1} + \frac{1}{t+1} \right) + \frac{1}{2} \log \left| \frac{t+1}{t-1} \right| + C \\[9pt] &= \frac{1}{2} \left( \frac{2t}{t^2-1} \right) + \frac{1}{2} \log \left| \frac{\sqrt{u} + 1}{\sqrt{u} - 1} \right| + C \\[9pt] &= \frac{1}{2} \left( \frac{2\sqrt{u}}{u-1} \right) + \frac{1}{2} \log \left| \frac{\sqrt{\frac{x+1}{x}} + 1}{\sqrt{\frac{x+1}{x}} - 1} \right| + C\\[9pt] &= \sqrt{x^2+x} + \frac{1}{2} \log \left| \frac{\sqrt{x+1} + \sqrt{x}}{\sqrt{x+1} - \sqrt{x}} \right| \\[9pt] &= \sqrt{x^2+x} + \frac{1}{2} \log \left( \sqrt{x+1} + \sqrt{x} \right)^2 + C\\[9pt] &= \sqrt{x^2+x} + \frac{1}{2} \log \left( 2 \sqrt{x^2+x} + 2x + 1 \right) + C. \end{align*}

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