Question

what is the integral of (x^2 *ln x)/2 . dx ??
please help :/

Answer 1

\[\int \frac{x^2\ln x}{2}dx\]\[=\frac{1}{2}\int\ln xd(\frac{x^3}{3})\]\[=\frac{1}{6}\int \ln xd(x^3)\]Then, by parts:\[=\frac{1}{6}(x^3(\ln x) - \int x^3d(\ln x))\]Should be alright now..

Answer 2

if you are not comfortable with the notation of d(x^3)
you can use the substitution, u=x^3

Answer 3

will the final answer be =2x-4x+c ??

Answer 4

how did u get that ?

Answer 5

lol see the attachment

Answer 6

aah, no
you took
dv = ln x dx
that does not give you v =1/x

Answer 7

follow what callisto suggested.

Answer 8

but how did he get (x33) ?

Answer 9

i mean did he get the derivative directly from the first step ?

Answer 10

3x^2 = d(x^3)
so,
x^2 = d(x^3/3)

if you are not comfortable with this way, try substituting u= x^3

Answer 11

ok mr hartnn i will try it now btw.. thank u for helping

Answer 12

is this correct ?? please say yes lol

Answer 13

\[\int udv = uv - \int vdu\]
u = ln(x)
v = \(\int x^2dx = \frac{x^3}{3}\)

Answer 14

ok i got it now thank u so much :D

Answer 15

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Answer 16

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