Question

Use integration by parts to prove the following reduction formulae:

Answer 1

\[\int\limits_{}^{}x^{n}\ln(x) dx=\frac{ x^{n+1} }{ (n+1)^{2} }[[1+(n+1)\ln(x)]+c\]

Answer 2

Integration by part: \[\large \int udv = uv-\int vdu\]

Answer 3

Let u be \(\large x^n\)
du=\(\large nx^{(n-1)}\)

dv=ln(x)\[\large v=\int ln(x)dx=x(ln(x)-1)+c\]

Answer 4

Apply the integration by part formula:\[\large \int x^nln(x)dx=ln(x)x^n-\int (x(ln(x)-1)nx^{n-1}dx\]

Answer 5

Um, should use du = ln(x) instead..

Answer 6

thats is what i did, i used u=lnx.

Answer 7

u=ln(x)
du=1/x dx
dv=x^n
v=integral of x^n dx = \(\LARGE \frac{x^{n+1}}{n+1}\)

Answer 8

yup that's what i had!

Answer 9

im down to:

Answer 10

So what's wrong with it? :(

Answer 11

\[\frac{ x^{n+1} }{ n+1 }lnx-\int\limits_{}^{}\frac{ x^{n-1} }{ n+1 }dx\]

Answer 12

is that right?

Answer 13

???

Answer 14

You forgot something in your integral

Answer 15

du is 1/x dx, not only dx

Answer 16

so the integral should be like \[\large \int \frac{x^{n+1}}{x(n+1)}dx\]

Answer 17

ok

Answer 18

and since 1/(n+1) is just a constant, move it out of the integral
\[\large \frac{1}{n+1}\int \frac{x^{n+1}}{x}dx\]

Answer 19

alrighty.

Answer 20

and that boils down to \[\large \frac{ln(x)x^{n+1}}{x^{n+1}}-\frac{1}{n+1}\int x^ndx\]

Answer 21

\[\large \frac{ln(x)x^{n+1}}{x^{n+1}}-\frac{1}{n+1}\frac{x^{n+1}}{n+1}+c\\\large \frac{ln(x)x^{n+1}}{x^{n+1}}-\frac{x^{n+1}}{(n+1)^2}+c\]

Answer 22

oups, the denominator of the first fraction is n+1, typo, my bad

Answer 23

that doesnt look like what the answer says it is tho in the quesiton stated? :S

Answer 24

\[\large \frac{ln(x)x^{n+1}}{n+1}-\frac{x^{n+1}}{(n+1)^2}+c\]

Answer 25

It will :)

Answer 26

is that the same thing?

Answer 27

They are, in the question, they just forged them into one single fraction

Answer 28

okay thanks!

Answer 29

np