Integral of $$$x \sqrt{x - 1}$$$

The calculator will find the integral/antiderivative of $$$x \sqrt{x - 1}$$$, with steps shown.

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Find $$$\int x \sqrt{x - 1}\, dx$$$.

Solution

Let $$$u=x - 1$$$.

Then $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral becomes

$${\color{red}{\int{x \sqrt{x - 1} d x}}} = {\color{red}{\int{\sqrt{u} \left(u + 1\right) d u}}}$$

Expand the expression:

$${\color{red}{\int{\sqrt{u} \left(u + 1\right) d u}}} = {\color{red}{\int{\left(u^{\frac{3}{2}} + \sqrt{u}\right)d u}}}$$

Integrate term by term:

$${\color{red}{\int{\left(u^{\frac{3}{2}} + \sqrt{u}\right)d u}}} = {\color{red}{\left(\int{\sqrt{u} d u} + \int{u^{\frac{3}{2}} d u}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{1}{2}$$$:

$$\int{u^{\frac{3}{2}} d u} + {\color{red}{\int{\sqrt{u} d u}}}=\int{u^{\frac{3}{2}} d u} + {\color{red}{\int{u^{\frac{1}{2}} d u}}}=\int{u^{\frac{3}{2}} d u} + {\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}=\int{u^{\frac{3}{2}} d u} + {\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{3}{2}$$$:

$$\frac{2 u^{\frac{3}{2}}}{3} + {\color{red}{\int{u^{\frac{3}{2}} d u}}}=\frac{2 u^{\frac{3}{2}}}{3} + {\color{red}{\frac{u^{1 + \frac{3}{2}}}{1 + \frac{3}{2}}}}=\frac{2 u^{\frac{3}{2}}}{3} + {\color{red}{\left(\frac{2 u^{\frac{5}{2}}}{5}\right)}}$$

Recall that $$$u=x - 1$$$:

$$\frac{2 {\color{red}{u}}^{\frac{3}{2}}}{3} + \frac{2 {\color{red}{u}}^{\frac{5}{2}}}{5} = \frac{2 {\color{red}{\left(x - 1\right)}}^{\frac{3}{2}}}{3} + \frac{2 {\color{red}{\left(x - 1\right)}}^{\frac{5}{2}}}{5}$$

Therefore,

$$\int{x \sqrt{x - 1} d x} = \frac{2 \left(x - 1\right)^{\frac{5}{2}}}{5} + \frac{2 \left(x - 1\right)^{\frac{3}{2}}}{3}$$

Simplify:

$$\int{x \sqrt{x - 1} d x} = \frac{2 \left(x - 1\right)^{\frac{3}{2}} \left(3 x + 2\right)}{15}$$

Add the constant of integration:

$$\int{x \sqrt{x - 1} d x} = \frac{2 \left(x - 1\right)^{\frac{3}{2}} \left(3 x + 2\right)}{15}+C$$

Answer

$$$\int x \sqrt{x - 1}\, dx = \frac{2 \left(x - 1\right)^{\frac{3}{2}} \left(3 x + 2\right)}{15} + C$$$A


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