Question

Integration using part formula mula1" title="\int\limits {x^nlogx} \, dx" alt="\int\limits {x^nlogx} \, dx" align="absmiddle" class="latex-formula">

Answer 1

Answer:

Integration of I= $\int {x^{n}logx \, dx$=$[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]$

Step-by-step explanation:

Given integral is I= $\int {x^{n}logx \, dx$

Take logx=t

$x=e^{t}$

$x^{n}=e^{nt}$

$\frac{1}{x} dx=dt$

$dx=xdt$

$dx=e^{t}dt$

I= $\int (e^{nt})(t)(e^{t})\, dt$

I= $\int (e^{(n+1)t})(t)\, dt$

Using integration by part,

$I= (t)\int [e^{(n+1)t}]\, dt-\int[\frac{d}{dt}{t}\times\int (e^{(n+1)t})]\\\\I= (t) [\frac{1}{n+1}e^{(n+1)t}]-\int[1\times\frac{1}{n+1}e^{(n+1)t}]\,dt\\\\I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]$

Writing in terms of x

I=$[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]$

I=$[\frac{logx}{n+1}e^{(n+1)logx}]-[\frac{1}{(n+1)^{2}}e^{(n+1)logx}]$

I=$[\frac{logx}{n+1}e^{logx^{(n+1)}}]-[\frac{1}{(n+1)^{2}}e^{logx^{(n+1)}}]$

I=$[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]$

Thus,

Integration of I= $\int {x^{n}logx \, dx$=$[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]$