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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} $

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  • (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,
Let $dv=(x^2+1)dx$,On integrating we get,
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,
$I=(log x)\frac{^3}{3}+x)-\frac{1}{3}\int x^2dx-\int dx.$
$\;\;\;=(log x)(\frac{x^3}{3}+x)-\frac{1}{3}.\frac{x^3}{3}-x+c.$
$\;\;\;=(log x)(\frac{x^3}{3}+x)-\frac{x^3}{9}-x+c.$


answered Feb 11, 2013 by sreemathi.v
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