Question

Integrate x^1/2.lnx dx

Answer 1

@Directrix @perl @ganeshie8 @jim_thompson5910 htye should know

Answer 2

@wio

Answer 3

let u=ln x
thus du/dx =1/x ie du=(1/x) dx
and then can u try to solve further?

Answer 4

is lnx consider as logarithmic or exponential?

Answer 5

i considered log here...
@wio is way better than me...can u help?

Answer 6

dv/dx = x^1/2 so, v = 2/3 x^3/2 right?

Answer 7

yep..:)

Answer 8

Is it: \[
\int x^{1/2}\ln (x)~dx
\]

Answer 9

you can do integration by parts

Answer 10

Thanks @shrutipande9. Yes, @wio. Noted, @perl

Answer 11

You'll use integration by parts
\[\large \int x^{1/2}*\ln(x)dx\]
\[\large \int \ln(x)*x^{1/2}dx\]

u = ln(x), du = (1/x)dx
dv = x^(1/2)dx, v = (2/3)x^(3/2)

\[\Large \int u dv = uv - \int v du\]
\[\large \int \ln(x)*x^{1/2}dx = \ln(x)*\left(\frac{2}{3}x^{3/2}\right)-\int \frac{2}{3}x^{3/2}\frac{1}{x}dx\]
I'll let you finish

Answer 12

oops, im stuck. D: @jim_thompson5910

Answer 13

I got 2/3 x ^3/2 lnx - 2/3 (x^1/2)

Answer 14

u = ln(x), du = dx/x
dv = x^(1/2)dx, v = (2/3)x^(3/2)

\[\Large \int u dv = uv - \int v du\]
\[\large \int \ln(x)*x^{1/2}dx = \ln(x)*\left(\frac{2}{3}x^{3/2}\right)-\int \frac{2}{3}x^{3/2}\frac{1}{x}dx\]
\[\large \int \ln(x)*x^{1/2}dx = \ln(x)*\left(\frac{2}{3}x^{3/2}\right)-\int \frac{2}{3}x^{3/2}x^{-1}dx\]
\[\large \int \ln(x)*x^{1/2}dx = \ln(x)*\left(\frac{2}{3}x^{3/2}\right)-\frac{2}{3}\int x^{3/2-1}dx\]
\[\large \int \ln(x)*x^{1/2}dx = \ln(x)*\left(\frac{2}{3}x^{3/2}\right)-\frac{2}{3}\int x^{1/2}dx\]
How about now?

Answer 15

exactly what i replied you before.

Answer 16

I see. You left off the integral sign, but that's ok

Answer 17

@jim_thompson5910 i am not that good in int...:/
was going right? sorry if i misleaded u @LXelle ..i tried..:)

Answer 18

so what is \[\Large \int x^{1/2} dx\] equal to?

Answer 19

@shrutipande9 It's okay. :)) You're doing fine.
@jim_thompson5910 2/3 x ^3/2 right?

Answer 20

correct

Answer 21

THANKS ALL! YOU GUYS ARE AMAZING! :)

Answer 22

u = ln(x), du = dx/x
dv = x^(1/2)dx, v = (2/3)x^(3/2)

\[\Large \int u dv = uv - \int v du\]
\[\large \int \ln(x)*x^{1/2}dx = \ln(x)*\left(\frac{2}{3}x^{3/2}\right)-\int \frac{2}{3}x^{3/2}\frac{1}{x}dx\]
\[\large \int \ln(x)*x^{1/2}dx = \ln(x)*\left(\frac{2}{3}x^{3/2}\right)-\int \frac{2}{3}x^{3/2}x^{-1}dx\]
\[\large \int \ln(x)*x^{1/2}dx = \ln(x)*\left(\frac{2}{3}x^{3/2}\right)-\frac{2}{3}\int x^{3/2-1}dx\]
\[\large \int \ln(x)*x^{1/2}dx = \ln(x)*\left(\frac{2}{3}x^{3/2}\right)-\frac{2}{3}\int x^{1/2}dx\]
\[\large \int \ln(x)*x^{1/2}dx = \ln(x)*\left(\frac{2}{3}x^{3/2}\right)-\frac{2}{3}*\left(\frac{2}{3}x^{3/2}\right)+C\]
\[\large \int \ln(x)*x^{1/2}dx = \frac{2}{3}x^{3/2}*\ln(x)-\frac{4}{9}*x^{3/2}+C\]
Optionally you can factor out (2/3)x^(3/2) and you'd get
\[\large \int \ln(x)*x^{1/2}dx = \frac{2}{3}x^{3/2}\left(\ln(x)-\frac{2}{3}\right)+C\]