# 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 $\sqrt x = u$. – Kaster Aug 9 '15 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. – Tucker Aug 9 '15 at 18:09
• @Kaster why is that inherently obvious, from the perspective of a beginner? – graydad Aug 9 '15 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. – graydad Aug 9 '15 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. – Tucker Aug 9 '15 at 18:19

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

• Simple but good... +1 – johannesvalks Aug 9 '15 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. – Shubham Marathia Aug 9 '15 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 $2\int (t^2-2)t^2$... – johannesvalks Aug 9 '15 at 18:19

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