Integrals — Class 12 Maths Solution

Let $I = \int {x{{\left( {\log x} \right)}^2}} dx = \int {{{\left( {\log x} \right)}^2} \cdot x{\kern 1pt} dx}$

$= {\left( {\log x} \right)^2}\int x dx - \int {\left( {\cfrac{d}{{dx}}{{\left( {\log x} \right)}^2} \cdot \int x dx} \right)} dx$

$= {\left( {\log x} \right)^2} \cdot \cfrac{{{x^2}}}{2} - \int {\left[ {\left( {2\log x} \right) \cdot \cfrac{1}{x}} \right]\left( {\cfrac{{{x^2}}}{2}} \right)} dx + C$

$= \cfrac{{{x^2}}}{2}{\left( {\log x} \right)^2} - \int {\left( {\log x} \right) \cdot x\,dx} + C$

$= \cfrac{{{x^2}}}{2}{\left( {\log x} \right)^2} - \left[ {\left( {\log \,x} \right) \cdot \cfrac{{{x^2}}}{2} - \int {\cfrac{1}{x} \cdot \cfrac{{{x^2}}}{2}} dx} \right] + C$

$= \cfrac{{{x^2}}}{2}{\left( {\log x} \right)^2} - \cfrac{{{x^2}}}{2}\log x + \cfrac{1}{2}\int {x\,dx} + C$

$= \cfrac{{{x^2}}}{2}{\left( {\log x} \right)^2} - \cfrac{{{x^2}}}{2}\log x + \cfrac{1}{2} \cdot \cfrac{{{x^2}}}{2} + C$

$= \cfrac{{{x^2}}}{2}{\left( {\log x} \right)^2} - \cfrac{{{x^2}}}{2}\log x + \cfrac{1}{4}{x^2} + C$