# 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 $u$.

• Obviously $\sqrt x = u$. Aug 9, 2015 at 18:07
• Why not distribute? The integral of the sum is the sum of the integrals and so the problem will require two uses of the power rule. Aug 9, 2015 at 18:09
• @Kaster why is that inherently obvious, from the perspective of a beginner? Aug 9, 2015 at 18:12
• @Kaster I do agree with your reasoning; this makes the substitution you proposed obvious. But a beginner might benefit from such an explanation before calling it obvious. Aug 9, 2015 at 18:17
• I would say that it is not obvious. I usually want to let $u$ be a function inside of another function like $e^{sin(x)}$. I would let $u$ be the inside of the exponential $sin(x)$. I better hope that there is a derivative of this thing floating around. Like maybe the problem were $\int cos(x)e^{sin(x)}dx$. Upon substitution the integral would look like $\int e^{u}du$ in the new variables. Aug 9, 2015 at 18:19

$$\int(x-1)\sqrt x \, dx = \int (x^{3/2} - x^{1/2})\, dx = \cdots$$

• Simple but good... +1 Aug 9, 2015 at 18:16

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 $C$.

Just change it to : $$x^{3/2} - x^{1/2}$$ and integrate it.

• @graydad Thanks for editing it for me. Aug 9, 2015 at 18:18

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

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 $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int (x-1)\sqrt xdx =\frac{2}{5}x^{5/2}-\frac{2}{3}x^{3/2}+c}}$$

$\bf{My\; Solution}$ 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 $$\displaystyle I = \int (x-1)\sqrt{x}dx = \frac{2}{5}x^\frac{5}{2}-\frac{2}{3}x^{\frac{3}{2}}+\mathcal{C}$$

• You miss a $t$ you get $2\int (t^2-2)t^2$... Aug 9, 2015 at 18:19

$$\int (x-1)\sqrt{x} \, \text{d} x=\int (x\sqrt{x}-\sqrt{x}) \, \text{d} x=\int (x\sqrt{x})\, \text{d} x-\int(\sqrt{x}) \, \text{d} x=$$ $$\int x\sqrt{x}\, \text{d} x-\int\sqrt{x} \, \text{d} x=\int x^{\frac{3}{2}}\, \text{d} x-\int x^{\frac{1}{2}} \, \text{d} x=\frac{1}{\frac{5}{2}}x^{\frac{3}{2}+1}-\int x^{\frac{1}{2}} \, \text{d} x=$$ $$\frac{2}{5}x^{\frac{5}{2}}-\int x^{\frac{1}{2}} \, \text{d} x=\frac{2x^{\frac{5}{2}}}{5}-\int x^{\frac{1}{2}} \, \text{d} x=$$ $$\frac{2x^{\frac{5}{2}}}{5}-\frac{1}{\frac{1}{2}+1}x^{\frac{1}{2}+1}=\frac{2x^{\frac{5}{2}}}{5}-\frac{2}{3}x^{\frac{3}{2}}=\frac{2x^{\frac{5}{2}}}{5}-\frac{2x^{\frac{3}{2}}}{3}+C$$