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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 $u$.

graydad
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Haim
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  • Obviously $\sqrt x = u$. – Kaster Aug 09 '15 at 18:07
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    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 09 '15 at 18:09
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    @Kaster why is that inherently obvious, from the perspective of a beginner? – graydad Aug 09 '15 at 18:12
  • @graydad from the prospective of the beginner, there are two terms in the relation – $x$ and $\sqrt x$. One can choose $u = x$, but even beginner should know that it won't lead to anything meaningful in terms of the problem. So, another choice that one should check immediately is the second term left - $\sqrt x$. – Kaster Aug 09 '15 at 18:16
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    @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 09 '15 at 18:17
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    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 09 '15 at 18:19

7 Answers7

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

Michael Hardy
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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 $C$.

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

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

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

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

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

Jan Eerland
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