Question Number 15235 by arnabpapu550@gmail.com last updated on 08/Jun/17
$$\int\left(\mathrm{log}\:\sqrt{\mathrm{x}\:}\:\right)^{\mathrm{2}} \mathrm{dx}=? \\ $$
Answered by liday last updated on 18/Jun/17
![∫(log(√x))^2 dx=(1/4)∫(logx)^2 dx=(1/4)[x(logx)^2 −∫x∙2logx∙(1/x)dx] =(1/4)[x(logx)^2 −2∫logxdx]=(1/4)[x(logx)^2 −2(xlogx−x)] =(x/4)[(logx)^2 −2logx+2]](Q16200.png)
$$\int\left({log}\sqrt{{x}}\right)^{\mathrm{2}} {dx}=\frac{\mathrm{1}}{\mathrm{4}}\int\left({logx}\right)^{\mathrm{2}} {dx}=\frac{\mathrm{1}}{\mathrm{4}}\left[{x}\left({logx}\right)^{\mathrm{2}} −\int{x}\centerdot\mathrm{2}{logx}\centerdot\frac{\mathrm{1}}{{x}}{dx}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[{x}\left({logx}\right)^{\mathrm{2}} −\mathrm{2}\int{logxdx}\right]=\frac{\mathrm{1}}{\mathrm{4}}\left[{x}\left({logx}\right)^{\mathrm{2}} −\mathrm{2}\left({xlogx}−{x}\right)\right] \\ $$$$=\frac{{x}}{\mathrm{4}}\left[\left({logx}\right)^{\mathrm{2}} −\mathrm{2}{logx}+\mathrm{2}\right] \\ $$
Commented by liday last updated on 18/Jun/17
$${and}+{c} \\ $$