Evaluating $\int x \sqrt{x^2 - x}\ dx$

Question

Problem is to integrate $ x \sqrt{x^2 - x}$.

My attempt:

I made it ready for a substitution $u = x^2 - x$

$$\begin{aligned} \int x \sqrt{x^2 -x}\ dx &= \int (2x-1)\sqrt{x^2 - x} \ dx - \int(x - 1)\sqrt{x^2 - x}\ dx\\& = \int \sqrt u\ du - \int(x - 1)\sqrt{x^2 - x}\ dx\\& = \frac23(x^2 -x)^{3/2} + C_1 - \int(x-1) \sqrt{x^2 -x}\ dx\end{aligned}$$ I don't know how to continue from here.

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Alternatively I tried this: $$\begin{aligned} \int x \sqrt{x^2 - x}\ dx &= \int x^2 \sqrt{1- \frac{1}{x}}dx\\ & \overset{1- \frac1x = t^2}{=} \int \frac{2t^2}{(1-t^2)^4}\ dt\\& \overset{t =\sin(\theta)}{=} \int \frac{2\sin^2(\theta) \cos(\theta)\ d\theta}{\cos^4(\theta)}\\& = \int 2 \tan(\theta)\tan(\theta) \sec(\theta) \ d\theta\\ & = \int 2\sqrt{\sec^2(\theta) - 1}\tan(\theta) \sec(\theta) \ d\theta\\& \overset{\sec(\theta) = u}{=} \int 2 \sqrt{u^2 - 1}\ du\\& = u \sqrt{u^2 - 1} - \ln|u + \sqrt{u^2- 1}| + C\\& = \sqrt{x^2 - x} - \ln|x + \sqrt{x^2 - x}| + C\end{aligned}$$

This method is very tedious. Is there any easy way to do the original integral?

Answer 1

Continue with the first approach \begin{aligned} I= \int x \sqrt{x^2 -x}\ dx=& \ \frac23(x^2 -x)^{3/2} - I+\int \sqrt{x^2 -x}\ dx\\ =&\ \frac13 (x^2 -x)^{3/2}+ \frac12\int \sqrt{x^2 -x}\ dx\\ \end{aligned} where $$\int \sqrt{x^2 -x}\ dx = \frac12(x-\frac12) \sqrt{x^2 -x}-\frac18\tanh^{-1}\frac{\sqrt{x^2 -x}}{x-\frac12} $$

Answer 2

$$I=\int x\sqrt{x^2-x}\, \mathrm{d}x$$ First of all, complete the square: $$I=\int x\sqrt{x^2-x}\, \mathrm{d}x=\int x\sqrt{\left(x-\frac{1}{2}\right)^2-\frac{1}{4}}\, \mathrm{d}x $$ Now let $x-1/2=u$: $$I=\frac{1}{2}\int \sqrt{u^2-\frac{1}{4}} \mathrm{d}u+\int u\sqrt{u^2-\frac{1}{4}}\mathrm{d}u$$ In the first integral, let $u=\dfrac{\sec \vartheta}{2}$; the second one is an immediate consequence of the chain rule.

Answer 3

$$ \begin{aligned} I & =\int x \sqrt{x^2-x} d x \\ & =\int\left(\frac{2 x-1}{2}+\frac{1}{2}\right) \sqrt{x^2-x} d x \\ & =\frac{1}{2} \int \sqrt{x^2-x} d\left(x^2-x\right)+\frac{1}{2} \int \sqrt{x^2-x} d x \\ & =\frac{1}{3}\left(x^2-x\right)^{\frac{3}{2}}+\frac{1}{2} \underbrace{\int \sqrt{\left(x-\frac{1}{2}\right)^2-\frac{1}{4}} d x}_{J} \end{aligned} $$ Let $x-\frac{1}{2}=\frac{1}{2} \sec \theta$ and using integration by parts, then $$ \begin{aligned} J & =\frac{1}{4} \int \tan ^2 \theta \sec \theta d \theta \\ & =\frac{1}{8}(\tan \theta \sec \theta-\ln |\tan \theta+\sec \theta \mid)+C \end{aligned} $$ Plugging back $x$ yields $$ I=\frac{1}{3}\left(x^2-x\right)^{\frac{3}{2}}+\frac{1}{16}\left[2(2 x-1) \sqrt{x^2-x}-\ln \left|2 \sqrt{x^2-x}+2 x-1\right|\right]+C $$

Answer 4

Substitute

$$t=\sqrt{x^2-x}-x \implies x = -\frac{t^2}{2t+1} \implies dx = -\frac{2t^2+2t}{(2t+1)^2} \, dt$$

and the resulting integrand is primed for partial fraction expansion.

$$\int x \sqrt{x^2-x} \, dx = 2 \int \frac{t^4(t+1)^2}{(2t+1)^4} \, dt$$

Answer 5

Hyperbolic function for $\int \sqrt{u^2-u}du$, using integration by parts yields $$ \begin{aligned} K & =u \sqrt{u^2-1}-\int \frac{u^2-1+1}{\sqrt{u^2-1}} d u \\ & =u \sqrt{u^2-1}-K-\int \frac{d u}{\sqrt{u^2-1}} \end{aligned} $$ Putting $y=\cosh u$ yields $$ K=\frac{1}{2}\left(u \sqrt{u^2-1}-\operatorname{arccosh} u\right)+C $$ Hence $$ I=\frac{1}{3}\left(x^2-x\right)^{\frac{3}{2}}+\frac{1}{16}\left[2(2 x-1) \sqrt{x^2-x}-\operatorname{arccosh}(2 x-1)\right] +C $$