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Question Number 74016 by mathmax by abdo last updated on 17/Nov/19

$$letg\left(x\right)=\frac{1}{x}{\int }_x^{2x+1}arctan\left(xt\right)dt$$$$find{lim}_{x\to 0}g\left(x\right)and{lim}_{x\to +\mathrm{\infty }}g\left(x\right).$$

Commented by mathmax by abdo last updated on 18/Nov/19

$$wehavexg\left(x\right)={\int }_x^{2x+1}arctan\left(xt\right)dt$$$${=}_{xt=u}{\int }_{x^2}^{x(2x+1)}arctan\left(u\right)\frac{du}{x}=\frac{1}{x}{\int }_{x^2}^{2x^2+x}arctan\left(u\right)du$$$$\Rightarrow x^{2}g\left(x\right){=}_{byparts}{\left[uarctan\left(u\right)\right]}_{x^2}^{2x^2+x}-{\int }_{x^2}^{2x^2+x}\frac{u}{1+u^2}du$$$$=(2x^2+x)arctan(2x^2+x)-x^{2}arctan\left(x^2\right)-{\left[\frac{1}{2}ln(1+u^2)\right]}_{x^2}^{2x^2+x}$$$$=(2x^2+x)arctan(2x^2+x)-x^{2}arctan\left(x^2\right)$$$$-\frac{1}{2}ln\left(\frac{1+{(2x^2+x)}^2}{1+x^4}\right)\Rightarrow$$$$g\left(x\right)=(2+\frac{1}{x})arctan(2x^2+x)-arctan\left(x^2\right)-\frac{1}{2x^2}ln\left(\frac{1+4x^4+4x^3+x^2}{1+x^4}\right)$$$$g\left(x\right)\sim \left(\frac{2x+1}{x}\right)(2x^2+x)-x^2-\frac{1}{2x^2}ln\left(\frac{4x^4+4x^3+x^2+1}{x^4+1}\right)(x\to 0)$$$${lim}_{x\to 0}g\left(x\right)=-\mathrm{\infty }$$$${lim}_{x\to +\mathrm{\infty }}g\left(x\right)=2\times \frac{\pi }{2}-\frac{\pi }{2}-0=\frac{\pi }{2}$$$$$$

Commented by mathmax by abdo last updated on 18/Nov/19

$$g\left(x\right)=(2+\frac{1}{x})arctan(2x^2+x)-\frac{arctan\left(x^2\right)}{x^2}-\frac{1}{2x^2}ln\left(\frac{4x^4+4x^3+x^2+1}{x^4+1}\right)$$

Commented by mind is power last updated on 18/Nov/19

$$\ln \left(\frac{1+{4\mathrm{x}}^4+{4\mathrm{x}}^3+{\mathrm{x}}^2}{{\mathrm{x}}^4+1}\right)=\ln (1+{\mathrm{x}}^2+{4\mathrm{x}}^3+{4\mathrm{x}}^4)-\ln (1+{\mathrm{x}}^4)$$$$\ln (1+{\mathrm{x}}^2+{4\mathrm{x}}^3+{4\mathrm{x}}^4)-\ln (1+{\mathrm{x}}^4)$$$$={\mathrm{x}}^2+0\left({\mathrm{x}}^2\right)\mathrm{x}\to 0$$$$$$

Commented by abdomathmax last updated on 18/Nov/19

$$thisquantityismultipliedby-\frac{1}{2x^2}sir....$$

Commented by mind is power last updated on 19/Nov/19

$$yeahsolim-\frac{1}{2x^2}ln\frac{(1+4x^4+4x^3+x^2)}{x^4+1}=-\frac{1}{2}$$

Answered by mind is power last updated on 18/Nov/19

$$\int \arctan \left(\mathrm{xt}\right)\mathrm{dt},\mathrm{u}=\mathrm{xt}$$$$=\frac{1}{\mathrm{x}}\int \arctan \left(\mathrm{u}\right)\mathrm{du}$$$$=\frac{1}{\mathrm{x}}\left[\mathrm{uarctab}\left(\mathrm{u}\right)\right]-\frac{1}{\mathrm{x}}\int \frac{\mathrm{u}}{1+{\mathrm{u}}^2}\mathrm{du}$$$$=\frac{1}{\mathrm{x}}\left[\mathrm{u}\mathrm{arcran}\left(\mathrm{u}\right)\right]-\frac{1}{2\mathrm{x}}.\ln (1+{\mathrm{u}}^2)$$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^2}.(\mathrm{x}(2\mathrm{x}+1)\arctan \left(\mathrm{x}(2\mathrm{x}+1)\right)-\frac{\ln (1+{\mathrm{x}}^2{(2\mathrm{x}+1)}^2)}{2}-{\mathrm{x}}^2\arctan \left({\mathrm{x}}^2\right)+\frac{\ln (1+{\mathrm{x}}^4)}{2}\}$$$$\underset{x\to 0}{\lim }\mathrm{g}\left(\mathrm{x}\right)=$$$$\ln (1+\mathrm{t})=\mathrm{t}+\mathrm{o}\left(\mathrm{t}\right)$$$$\arctan \left(\mathrm{t}\right)=\mathrm{t}+\mathrm{o}\left(\mathrm{t}\right)$$$$\Rightarrow \underset{x\to 0}{\lim }\mathrm{g}\left(\mathrm{x}\right)=(1-\frac{1}{2}-0+0)=\frac{1}{2}$$$$\underset{x\to +\mathrm{\infty }}{\lim g}\left(\mathrm{x}\right)=\pi -\frac{\pi }{2}=\frac{\pi }{2}$$$$$$

Commented by abdomathmax last updated on 18/Nov/19

$$thankyousir.$$

Commented by mind is power last updated on 18/Nov/19

$${\hat{\mathrm{y}}}^{\prime }\mathrm{re}\mathrm{welcom}$$

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