How to integrate $\int (x-1)\sqrt{x} \, \text{d}x$ How do I find this integral:
$$\int (x-1)\sqrt{x} \, \text{d} x$$
I thought to use use substitution, but am not sure what I should use as $u$. 
 A: You can probably anti-differentiate it easily enough. $$(x-1)\sqrt{x} = x^{3/2}-x^{1/2}$$ Now just use the power rule in reverse and add a constant $C$.
A: Just change it to : $$x^{3/2} - x^{1/2}$$ and integrate it.
A: $$\int { \left( x-1 \right) \sqrt { x } dx } =\int { \left( { x }^{ \frac { 3 }{ 2 }  }-{ x }^{ \frac { 1 }{ 2 }  } \right) dx= } \int { { x }^{ \frac { 3 }{ 2 }  } } dx-\int { { x }^{ \frac { 1 }{ 2 }  } } dx=\\ =\frac { 2 }{ 5 } { x }^{ \frac { 5 }{ 2 }  }-\frac { 2 }{ 3 } x^{ \frac { 3 }{ 2 }  }+C$$
A: Notice, the following formula $$\int x^n dx=\frac{x^{n+1}}{n+1}+c$$ Now, we have $$=\int (x-1)\sqrt xdx $$ $$=\int (x-1)x^{1/2}dx $$ $$=\int (x^{3/2}-x^{1/2})dx $$ $$=\frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}-\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c $$
$$=\frac{2}{5}x^{5/2}-\frac{2}{3}x^{3/2}+c $$
Hence, we have $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int (x-1)\sqrt xdx =\frac{2}{5}x^{5/2}-\frac{2}{3}x^{3/2}+c}}$$
A: $$
\int(x-1)\sqrt x \, dx = \int (x^{3/2} - x^{1/2})\, dx = \cdots
$$
A: $\bf{My\; Solution}$ Given $\displaystyle \int (x-1)\sqrt{x}dx\;,$ Let $x=t^2\;,$ Then $dx = 2tdt$
So Integral $$\displaystyle I = 2\int (t^2-1)t^2dt = 2\int t^4dt-2\int t^2dt = \frac{2}{5}t^5-\frac{2}{3}t^3+\mathcal{C}$$
So $$\displaystyle I = \int (x-1)\sqrt{x}dx = \frac{2}{5}x^\frac{5}{2}-\frac{2}{3}x^{\frac{3}{2}}+\mathcal{C}$$
A: $$\int (x-1)\sqrt{x} \, \text{d} x=\int (x\sqrt{x}-\sqrt{x}) \, \text{d} x=\int (x\sqrt{x})\, \text{d} x-\int(\sqrt{x}) \, \text{d} x=$$
$$\int x\sqrt{x}\, \text{d} x-\int\sqrt{x} \, \text{d} x=\int x^{\frac{3}{2}}\, \text{d} x-\int x^{\frac{1}{2}} \, \text{d} x=\frac{1}{\frac{5}{2}}x^{\frac{3}{2}+1}-\int x^{\frac{1}{2}} \, \text{d} x=$$
$$\frac{2}{5}x^{\frac{5}{2}}-\int x^{\frac{1}{2}} \, \text{d} x=\frac{2x^{\frac{5}{2}}}{5}-\int x^{\frac{1}{2}} \, \text{d} x=$$
$$\frac{2x^{\frac{5}{2}}}{5}-\frac{1}{\frac{1}{2}+1}x^{\frac{1}{2}+1}=\frac{2x^{\frac{5}{2}}}{5}-\frac{2}{3}x^{\frac{3}{2}}=\frac{2x^{\frac{5}{2}}}{5}-\frac{2x^{\frac{3}{2}}}{3}+C$$
