Respuesta :
Answer:
[tex]ln|x|(\frac{x^{n+1}}{n+1})-\frac{x^{n+1}}{(n+1)^2}+C,n\neq -1[/tex]
Step-by-step explanation:
We have been given an indefinite integral [tex]\int \:x^n\:.\:ln\:|x|dx[/tex]. We are asked to integrate it.
We will use integration by pats to solve our given problem.
[tex]\int udvdx=uv-\int vdu[/tex]
Let [tex]u=ln|x|[/tex] and [tex]v'=x^n[/tex].
Now, we need to find du and v using above values.
[tex]\frac{du}{dx}=\frac{1}{x}[/tex]
[tex]du=\frac{1}{x}dx[/tex]
[tex]v'=x^n[/tex]
[tex]v=\frac{x^{n+1}}{n+1}[/tex]
Substitute these values in integration by parts formula:
[tex]\int \:x^n\:.\:ln\:|x|dx=ln|x|(\frac{x^{n+1}}{n+1})-\int \frac{x^{n+1}}{n+1}*\frac{1}{x}dx[/tex]
[tex]\int \:x^n\:.\:ln\:|x|dx=ln|x|(\frac{x^{n+1}}{n+1})-\int \frac{x^{n+1}}{n+1}*x^{-1}dx[/tex]
[tex]\int \:x^n\:.\:ln\:|x|dx=ln|x|(\frac{x^{n+1}}{n+1})-\frac{1}{n+1}\int x^{n+1-1}dx[/tex]
[tex]\int \:x^n\:.\:ln\:|x|dx=ln|x|(\frac{x^{n+1}}{n+1})-\frac{1}{n+1}\int x^{n}dx[/tex]
[tex]\int \:x^n\:.\:ln\:|x|dx=ln|x|(\frac{x^{n+1}}{n+1})-\frac{1}{n+1}*\frac{x^{n+1}}{n+1}+C[/tex]
[tex]\int \:x^n\:.\:ln\:|x|dx=ln|x|(\frac{x^{n+1}}{n+1})-\frac{x^{n+1}}{(n+1)^2}+C,n\neq -1[/tex]
Therefore, our required integral would be [tex]ln|x|(\frac{x^{n+1}}{n+1})-\frac{x^{n+1}}{(n+1)^2}+C,n\neq -1[/tex].
