# partial integration problem

• Jun 10th 2009, 06:44 AM
coobe
partial integration problem
Hello!

i have difficulties sovling the following equation:

$\displaystyle \int \frac{ln(x)} {\sqrt{x}} dx$

the actual integration part isnt the problem:

$\displaystyle ln(x)*2\sqrt{x} \int x^\frac{-2}{3} dx$

simplifying the integral: $\displaystyle \int x^\frac{-3}{2}$

and after integrating im here:

$\displaystyle ln(x)*2\sqrt{x} * (-\frac{2}{\sqrt(x)})+C$

the solution is supposed to be $\displaystyle 2*ln(x)*\sqrt(x)-4\sqrt(x)+C$

i dont know how to get this result :(
• Jun 10th 2009, 06:52 AM
TheEmptySet
Quote:

Originally Posted by coobe
Hello!

i have difficulties sovling the following equation:

$\displaystyle \int \frac{ln(x)} {\sqrt{x}} dx$

the actual integration part isnt the problem:

$\displaystyle ln(x)*2\sqrt{x} \int x^\frac{-2}{3} dx$

simplifying the integral: $\displaystyle \int x^\frac{-3}{2}$

and after integrating im here:

$\displaystyle ln(x)*2\sqrt{x} * (-\frac{2}{\sqrt(x)})+C$

the solution is supposed to be $\displaystyle 2*ln(x)*\sqrt(x)-4\sqrt(x)+C$

i dont know how to get this result :(

$\displaystyle \int \frac{ln(x)} {\sqrt{x}} dx$

so let

$\displaystyle u=\ln(x) \implies du =\frac{1}{x}dx$
$\displaystyle dv=\frac{1}{\sqrt{x}}dx=x^{-1/2}dx \implies v =2\sqrt{x}dx$

$\displaystyle \int \frac{ln(x)} {\sqrt{x}} dx=2\ln(x) \sqrt{x}-2\int x^{-1/2}dx$

$\displaystyle \int \frac{ln(x)} {\sqrt{x}} dx=2\ln(x) \sqrt{x}-4\sqrt{x}+c$
• Jun 10th 2009, 06:57 AM
Amer
Quote:

Originally Posted by coobe
Hello!

i have difficulties sovling the following equation:

$\displaystyle \int \frac{ln(x)} {\sqrt{x}} dx$

the actual integration part isnt the problem:

$\displaystyle ln(x)*2\sqrt{x} \int x^\frac{-2}{3} dx$

simplifying the integral: $\displaystyle \int x^\frac{-3}{2}$

and after integrating im here:

$\displaystyle ln(x)*2\sqrt{x} * (-\frac{2}{\sqrt(x)})+C$

the solution is supposed to be $\displaystyle 2*ln(x)*\sqrt(x)-4\sqrt(x)+C$

i dont know how to get this result :(

let

$\displaystyle lnx=u...x=e^u...dx=e^udu...x^{\frac{1}{2}}=e^{\fra c{u}{2}}$

$\displaystyle \int \frac{u(e^u)}{e^{\frac{u}{2}}} du$$\displaystyle =\int u(e^{u-\frac{u}{2}}) du$

$\displaystyle \int ue^{\frac{u}{2}} du =$

by parts$\displaystyle dy=e^{\frac{u}{2}} ...y=2e^{\frac{u}{2}}...and...t=u...dt=du$

$\displaystyle 2ue^{\frac{u}{2}}-2\int e^{\frac{u}{2}} du$

$\displaystyle 2ue^{\frac{u}{2}} -2(2)e^{\frac{u}{2}} + c$

sub $\displaystyle u=ln(x).......e^{\frac{u}{2}}=\sqrt{x}$

someone is faster than me lol
• Jun 10th 2009, 07:12 AM
Soroban
Hello, coobe!

I'm not sure what you did . . .

Quote:

$\displaystyle \int \frac{\ln x} {\sqrt{x}}\,dx$
We have: .$\displaystyle I \;=\;\int\ln x\left(x^{-\frac{1}{2}}\,dx\right)$

Integrate by parts: .$\displaystyle \begin{array}{ccccccc} u &=&\ln x & & dv &=& x^{-\frac{1}{2}}dx \\ du &=& \frac{dx}{x} & & v &=& 2x^{\frac{1}{2}} \end{array}$

Then: .$\displaystyle I \;\;=\;\;2\!\cdot\!x^{\frac{1}{2}}\!\cdot\!\ln x - \int\left(2x^{\frac{1}{3}}\right)\left(\frac{dx}{x }\right)$

. . . . . $\displaystyle =\;\;2\!\cdot\!x^{\frac{1}{2}}\!\cdot\!\ln x - 2\int x^{-\frac{1}{2}}\,dx$

. . . . . $\displaystyle = \;\;2\cdot\ln x\cdot\!x^{\frac{1}{2}} - 4x^{\frac{1}{2}} + C$

. . . . . $\displaystyle = \;\;2\cdot\ln x\!\cdot\!\sqrt{x} - 4\sqrt{x} + C$

• Jun 10th 2009, 09:22 AM
coobe
d'oh.... i had a wrong formula on my Formula Sheet... took me so much time to figure that out :)

thanks everybody, i simply had a wrong formula for partial integrating...

good thing i noticed that before the exam