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Question Number 35682 by prof Abdo imad last updated on 22/May/18

$$letF\left(x\right)={\int }_{x+1}^{x^2+1}arctan(1+t)dt$$$$1)calculate\frac{\partial F}{\partial x}\left(x\right)$$$$2)find{lim}_{x\to 0}F\left(x\right).$$

Commented by tanmay.chaudhury50@gmail.com last updated on 22/May/18

Commented by abdo mathsup 649 cc last updated on 23/May/18

$$1)wehave\frac{\partial F}{\partial x}\left(x\right)={(x^2+1)}^{{}^{\prime }}arctan(2+x^2)$$$$-{(x+1)}^{{}^{\prime }}arctan(2+x)$$$$=2xarctan(3+x^2)-arctan(2+x)$$$$2)\mathrm{\exists }\xi \in ]x+1,x^2+1[/$$$$F\left(x\right)=arctan(1+\xi ){\int }_{x+1}^{x^2+1}dt$$$$=(x^2-x)arctan(1+\xi )\Rightarrow {lim}_{x\to 0}F\left(x\right)$$$$=0\times arctan\left(2\right)=0.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/May/18

$$\int {tan}^{-1}(1+t)dt$$$$={ttan}^{-1}(1+t)-\int \frac{1}{1+{(1+t)}^2}\times tdt$$$$=t\times {tan}^{-1}(t+1)-\int \frac{t}{t^2+2t+2}dt$$$$=t\times {tan}^{-1}(t+1)-\frac{1}{2}\int \frac{2t+2-2}{t^2+2t+2}dt$$$$=t\times {tan}^{-1}(t+1)-\frac{1}{2}\int \frac{2t+2}{t^2+2t+2}+\int \frac{dt}{1+{(t+1)}^2}$$$$=t\times {tan}^{-1}(t+1)-\frac{1}{2}\times ln(t^2+2t+2)+{tan}^{-1}(t+1$$$$nowputtingtheupperlimitandlowerlimit$$$$up=(x^2+1){tan}^{-1}(x^2+2)-\frac{1}{2}\times ln(x^4+2^{}{x}^2+1+$$$$2x^2+2+2)+{tan}^{-1}(x^2+2)$$$$$$$$lw=(x+1)\times {tan}^{-1}(x+2)-\frac{1}{2}\times ln(x^2+2x+1+$$$$2x+2+2)+{tan}^{-1}(x+2)$$$$requiredresult=\frac{\partial }{\partial x}(up-lw)$$$$pageareanotsufficient...$$

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