Answer:
Integration of I= [tex]\int {x^{n}logx \, dx[/tex]=[tex][\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}][/tex]
Step-by-step explanation:
Given integral is I= [tex]\int {x^{n}logx \, dx[/tex]
Take logx=t
[tex]x=e^{t}[/tex]
[tex]x^{n}=e^{nt}[/tex]
[tex]\frac{1}{x} dx=dt[/tex]
[tex]dx=xdt[/tex]
[tex]dx=e^{t}dt[/tex]
I= [tex]\int (e^{nt})(t)(e^{t})\, dt[/tex]
I= [tex]\int (e^{(n+1)t})(t)\, dt[/tex]
Using integration by part,
[tex]I= (t)\int [e^{(n+1)t}]\, dt-\int[\frac{d}{dt}{t}\times\int (e^{(n+1)t})]\\\\I= (t) [\frac{1}{n+1}e^{(n+1)t}]-\int[1\times\frac{1}{n+1}e^{(n+1)t}]\,dt\\\\I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}][/tex]
Writing in terms of x
I=[tex][\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}][/tex]
I=[tex][\frac{logx}{n+1}e^{(n+1)logx}]-[\frac{1}{(n+1)^{2}}e^{(n+1)logx}][/tex]
I=[tex][\frac{logx}{n+1}e^{logx^{(n+1)}}]-[\frac{1}{(n+1)^{2}}e^{logx^{(n+1)}}][/tex]
I=[tex][\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}][/tex]
Thus,
Integration of I= [tex]\int {x^{n}logx \, dx[/tex]=[tex][\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}][/tex]
