Question

Use integration by parts to prove the following reduction reduction formulae.

Answer 1

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

Answer 2

i know ill be using \[\int\limits_{}^{}udv=uv-\int\limits_{}^{}v du\]

Answer 3

and so far i've let u=x^(n+1)/(n+1)^2 so du=x^n/n+1 and i've let dv=dx so v=x

Answer 4

Is that "a" on the right side?

Answer 5

no it's x^n

Answer 6

oh it should be n+1 not n+a

Answer 7

You should choose dv as x^n

Answer 8

no a's in the equation at all sorry.

Answer 9

and should u be what?

Answer 10

\[\int x^n \ln x dx=\frac{x^{n+1}}{n+1}\times \ln x -\int \frac d {dx} \ln x\times \frac{x^n}{n+1} dx\]
\[\frac{x^{n+1}}{n+1}\times \ln x -\int \frac 1 {x} \times \frac{x^n}{n+1} dx\]
\[\frac{x^{n+1}}{n+1}\times \ln x -\int \frac{x^{n-1}}{n+1} dx\]
Can you solve from hee?

Answer 11

so i let u=ln x so du=1/x and i let dv = x^n so v=nx^(n-1) ??

Answer 12

*here?

Answer 13

was my previous statement right?

Answer 14

and im having problem solving the integral part.

Answer 15

use tabular integration

Answer 16

saimoulis meathod is easy tabular integration is much eaiser