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Q 15 Integrate the functions $( x^2 + 1 ) \log x$

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Consider $( x^2 + 1 ) \log x$

So, we have then: $I = \int (x^2+1) \log x dx = \int x^2 \log x dx +\int \log x dx$

Let us take $I = I_{1} +I_{2}$ ....................(1)

Where, $I_{1} = \int x^2\log x dx$ and $I_{2} = \int \log x dx$

So, $I_{1} = \int x^2\log x dx$

After taking $\log x$ as a first function and $x^2$ as second function and integrating by parts, we get

$I = \log x \int x^2dx -\int \left \{ \left ( \frac{d}{dx} \log x \right )\int x^2.dx \right \}dx$

$= \log x .\frac{x^3}{3} - \int \frac{1}{x}.\frac{x^3}{3} dx$

$= \log x .\frac{x^3}{3} - \frac{x^3}{9} +C_{1}$ ....................(2)

$I_{2} = \int \log x dx$

After taking $\log x$ as a first function and $1$ as second function and integrating by parts, we get

$I_{2} = \log x \int 1.dx - \int \left \{ \left ( \frac{d}{dx}\log x \right ) \int 1.dx \right \}dx$

$= \log x .x -\int \frac{1}{x}. xdx$

$= x\log x -\int 1 dx$

$= x\log x -x +C_{2}$ ................(3)

Now, using the two equations (2) and (3) in (1) we get,

$I = \frac{x^3}{3}\log x -\frac{x^3}{9} +C_{1} +x\log x - x +C_{2}$

$= \frac{x^3}{3}\log x -\frac{x^3}{9} +x\log x - x +(C_{1}+C_{2})$

$=\left ( \frac{x^3}{3}+x \right ) \log x -\frac{x^3}{9} -x+C$

Posted by

Divya Prakash Singh

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