Question

Integrate the function $(x^2+1)\log x$

Answer 1

Let $I=\int (x^2+1)\log xdx=\int x^2\log xdx+\int \log xdx$
Let $I=I_1+I_2$ ........... (1)
Where, $I_1=\int x^2\log xdx$ and $I_2=\int \log xdx$
$I_1=\int x^2\log xdx$
Taking $\log x$ as first function and $x^2$ as second function and integrating by parts, we obtain
$I_1=\log x\int x^{2}dx-\int \{\left(\frac{d}{dx}\log x\right)\int x^{2}dx\}dx$
$=\log x\cdot \frac{x^3}{3}-\int \frac{1}{x}\cdot \frac{x^3}{3}dx$
$=\frac{x^3}{3}\log x-\frac{1}{3}(\int x^{2}dx)$
$=\frac{x^3}{3}\log x-\frac{x^3}{9}+C_1$ .......... (2)
$I_2=\int \log xdx$
Taking $\log x$ as first function and $1$ as second function and integrating by part $x$, we obtain
$I_1=\log x\int 1\cdot dx-\int \{\left(\frac{d}{dx}\log x\right)\int 1\cdot dx\}$
$=\log x\cdot x-\int \frac{1}{x}\cdot xdx$
$=x\log x-\int 1dx$
$=x\log x-x+C_2$ ...... (3)
Using equation (2) and (3) in (1), we obtain
$I=\frac{x^3}{3}\log x-\frac{x^3}{9}+C_1+x\log x-x+C_2$
$=\frac{x^3}{3}\log x-\frac{x^3}{9}+x\log x-x+(C_1+C_2)$
$=(\frac{x^3}{3}+x)\log x-\frac{x^3}{9}-x+C$