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

$\begin{array}{1 1}\log x (x+\large \frac{x^3}{3})-\large\frac{x^3}{9}-x+c \\ \log x (x-\large \frac{x^3}{3})+\large\frac{x^3}{9}-x+c \\ \log x (x+\large \frac{x^4}{3})-\large\frac{x^4}{9}-x+c \\ \log x (x+\large \frac{x^2}{3})-\large\frac{x^2}{4}-x+c \end{array}$

Toolbox:
• (i)When there are two functions u and v and if they are of the form $\int u dv,$then we can solve it by the method of integration by parts$\int udv=uv-\int vdu$
• (ii)$\frac{d}{dx}(log x)=\frac{1}{x}.$
• (iii)$\int x^ndx=\frac{x^{n+1}}{n+1}+c.$
Given $I=\int (x^2+1)log x.$

Clearly the given integral function is of the form $\int u dv$

Let u=log x.On differentiating with respect to x,

$du=\frac{1}{x}dx.$

Let $dv=(x^2+1)dx$,On integrating we get,

$v=\frac{x^3}{3}+x.$

Now substituting for u,v,du and dv we get,

$\int (x^2+1)log x=(log x)(\frac{x^3}{3}+x)-\int(\frac{x^3}{3}+x).\frac{1}{x}dx.$

On cancelling the common terms,

$I=(log x)\frac{^3}{3}+x)-\frac{1}{3}\int x^2dx-\int dx.$

$\;\;\;=(log x)(\frac{x^3}{3}+x)-\frac{1}{3}.\frac{x^3}{3}-x+c.$

$\;\;\;=(log x)(\frac{x^3}{3}+x)-\frac{x^3}{9}-x+c.$