Question

Integrate the function: $(x^2+1)logx$

Answer 1

The correct answer is: $(\frac{x^3}{3}+x)logx-\frac{x^3}{9}-x+C$
Let $I=∫(x^2+1)logx\, dx=∫x^2 logx\, dx+∫logx\, dx$
Let $I=I_1+I_2...(1)$
Where,$I_1=∫x^2logx\, dx\,\, and \,\,I_2=∫logx dx$
$I_1=∫x^2logx\, dx$
Taking log $x$ as first function and $x^2$ as second function and integrating by parts,we
obtain
$I_1=logx-∫x^2dx-∫[{(\frac{d}{dx}logx)∫x^2dx}]dx$
$=logx.\frac{x^3}{3}-∫\frac{1}{x}.\frac{x^3}{3}dx$
$=\frac{x^3}{3}logx-\frac{1}{3}(∫x^2dx)$
$=\frac{x^3}{3}logx-\frac{x^3}{9}+C1...(2)$
$I_2=∫logx dx$
Taking log $x$ as first function and 1 as second function and integrating by parts,we
obtain
$I_2=logx∫1.dx-∫{(\frac{d}{dx}logx)∫1.dx}$
$=logx.x-∫\frac{1}{x}.xdx$
$=xlogx-∫1dx$
$=xlogx-x+C2...(3)$
Using equations(2)and(3)in(1),we obtain
$I=\frac{x^3}{3}logx-\frac{x^3}{9}+C_1+xlogx-x+C_2$
$=\frac{x^3}{3}logx-\frac{x^3}{9}+xlogx-x+(C_1+C_2)$
$=(\frac{x^3}{3}+x)logx-\frac{x^3}{9}-x+C$