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
\(I=\frac{x^3}{3}logx-\frac{x^3}{9}+C_1+xlogx-x+C_2\)
\(=\frac{x^3}{3}logx-\frac{x^3}{9}+xlogx-x+(C_1+C_2)\)
\(=(\frac{x^3}{3}+x)logx-\frac{x^3}{9}-x+C\)