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$$\int \sqrt{x}\log_2(x) \, dx$$

Integral of Square root of x times log base 2 of x dx.

Michael Hardy
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  • Since you know the method, what did you try before asking the question? Did you try to integrate the square, and then derive the log? – mvggz Oct 15 '14 at 22:27
  • make log base 2 of x--> $ln(x)/(ln(2)$, $u = ln(x)$, $v' = \sqrt (x)$ so I get $$2/3ln(x)x^1.5 - 2/3(\sqrt(x))/(ln(2))$$ –  Oct 15 '14 at 22:29

2 Answers2

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First lets change the base of the logarithm $$\log_2(x)=\frac{\log_e(x)}{\log_e(2)}=\frac{\ln(x)}{\ln(2)}$$ So your integral can be reformulated as $$\int\sqrt{x}\log_2(x)\,dx=\int\sqrt{x}\frac{\ln(x)}{\ln(2)}\,dx=\frac{1}{\ln(2)}\int\sqrt{x}\ln(x)\,dx$$ Integrate by parts $$\frac{1}{\ln(2)}\int\sqrt{x}\ln(x)\,dx=\frac{1}{\ln(2)}[\frac{2}{3}x^{3/2}\ln(x)-\frac{2}{3}\int x^{3/2}\cdot\frac{1}{x}\,dx]=\frac{1}{\ln(2)}[\frac{2}{3}x^{3/2}\ln(x)-\frac{4}{9}x^{3/2}+c]$$

Arian
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  • could I also get a hint for the integral of x(ln(x)) ? My idea was to set u as ln(x) and dv as x. Is there a better way? –  Oct 16 '14 at 01:42
  • @Ben:I think that is a good way to go because it is easier to integrate a function that is a power of $x$ then integrate $\ln(x)$. – Arian Oct 17 '14 at 20:39
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If $u=\log_2 x$ then $du= \dfrac {dx} {x\log_e 2}$. If $dv=\sqrt{x}\,dx$ then $v = \dfrac 2 3 x^{3/2}$. So you get \begin{align} & \int u\,dv=vu-\int v\, du = \frac 2 3 x^{3/2}\log_2 x - \int\frac 2 3 x^{3/2}\frac{dx}{x\log_e 2} \\[8pt] = {} & \frac 2 3 x^{3/2}\log_2 x - \frac{2}{3\log_e 2} \int x^{3/2}\frac {dx} x \\[8pt] = {} & \frac 2 3 x^{3/2}\log_2 x - \frac{2}{3\log_e 2} \int x^{1/2}\, dx= \cdots \end{align} Here I used the fact that $x^{3/2}\dfrac 1 x= x^{1/2}$. Strangely, I have found that simple point to be a place where students get stuck after getting everything else right.

Michael Hardy
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  • could I also get a hint for the integral of x(ln(x)) ? My idea was to set u as ln(x) and dv as x. Is there a better way? –  Oct 16 '14 at 01:42
  • $dv$ would be $x,dx$, not just $x$. Other than that, that's the way to do it. – Michael Hardy Oct 16 '14 at 02:13