Alright, let's break down how to compute the indefinite integral (\int \sqrt{ax^2 - a^2x} , dx) step by step. Completing the square is our key technique here, since we're dealing with a square root of a quadratic expression.

Step 1: Simplify the integrand by completing the square #

First, factor out the coefficient of (x^2) from the quadratic inside the square root:

ax² - a²x = a(x² - ax)

Next, complete the square for (x^2 - ax):

x² - ax = \left(x - \frac{a}{2}\right)^2 - \left(\frac{a}{2}\right)^2

Substitute this back into the integrand to rewrite it in a more manageable form:

\sqrt{ax^2 - a^2x} = \sqrt{a\left( \left(x - \frac{a}{2}\right)^2 - \left(\frac{a}{2}\right)^2 \right)} = \sqrt{a} \cdot \sqrt{\left(x - \frac{a}{2}\right)^2 - \left(\frac{a}{2}\right)^2}

Note: We assume (a > 0) here—if (a < 0), the expression inside the square root would be negative for all real (x), making the integral undefined over the real numbers.

Step 2: Substitute to reduce to a standard integral #

Let (u = x - \frac{a}{2}), so (du = dx). This transforms our integral into a standard form that we can use a known formula for:

\int \sqrt{ax^2 - a^2x} \, dx = \sqrt{a} \int \sqrt{u^2 - \left(\frac{a}{2}\right)^2} \, du

This matches the integral pattern (\int \sqrt{u^2 - b^2} , du), where (b = \frac{a}{2}).

Step 3: Apply the standard integral formula #

The formula for (\int \sqrt{u^2 - b^2} , du) is:

\frac{u}{2}\sqrt{u^2 - b^2} - \frac{b^2}{2} \ln\left| u + \sqrt{u^2 - b^2} \right| + C

Now substitute back (u = x - \frac{a}{2}) and (b = \frac{a}{2}):

First term calculation:

\frac{1}{2}\left(x - \frac{a}{2}\right)\sqrt{\left(x - \frac{a}{2}\right)^2 - \left(\frac{a}{2}\right)^2} = \frac{2x - a}{4}\sqrt{x^2 - ax}

Multiply by (\sqrt{a}) to get the first part of our result:

\frac{2x - a}{4}\sqrt{a(x^2 - ax)} = \frac{2x - a}{4}\sqrt{ax^2 - a^2x}

Second term calculation:

- \frac{\left(\frac{a}{2}\right)^2}{2} \ln\left| \left(x - \frac{a}{2}\right) + \sqrt{\left(x - \frac{a}{2}\right)^2 - \left(\frac{a}{2}\right)^2} \right| = - \frac{a^2}{8} \ln\left| x - \frac{a}{2} + \sqrt{x^2 - ax} \right|

Multiply by (\sqrt{a}) to get the second part:

- \frac{a^2\sqrt{a}}{8} \ln\left| x - \frac{a}{2} + \sqrt{x^2 - ax} \right| = - \frac{a^{5/2}}{8} \ln\left| x - \frac{a}{2} + \sqrt{x^2 - ax} \right|

Step 4: Combine terms for the final result #

Putting everything together, the indefinite integral is:

\int \sqrt{ax^2 - a^2x} \, dx = \frac{(2x - a)\sqrt{ax^2 - a^2x}}{4} - \frac{a^{5/2}}{8} \ln\left| x - \frac{a}{2} + \sqrt{x^2 - ax} \right| + C

If you prefer, you can rewrite (a^{5/2}) as (a^2\sqrt{a}) for clarity—both forms are equivalent.

As a quick check, if you differentiate this result with respect to (x), you should end up back with the original integrand (\sqrt{ax^2 - a^2x}), which confirms we got it right.

内容的提问来源于stack exchange，提问作者Nehal Samee