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Question Number 68001 by mathmax by abdo last updated on 03/Sep/19

$$letF\left(x\right)={\int }_{2x}^{x^2+1}\frac{e^{-xt}}{x+2t}dtcalculateF{}^{{}^{\prime }}\left(x\right)$$

Commented by mathmax by abdo last updated on 06/Sep/19

$$F\left(x\right)={\int }_{2x}^{x^2+1}\frac{e^{-xt}}{x+2t}dtwehaveu\left(x\right)=2x,v\left(x\right)=x^2+1,g(x,t)=\frac{e^{-xt}}{x+2t}$$$$weapllytbeformula\Rightarrow F{}^{{}^{\prime }}\left(x\right)={\int }_{u\left(x\right)}^{v\left(x\right)}\frac{\partial g}{\partial x}(x,t)dt+v^{{}^{\prime }}g(x,v)$$$$-u^{{}^{\prime }}g(x,u)={\int }_{2x}^{x^2+1}\frac{\partial g}{\partial x}(x,t)dt+2xg(x,x^2+1)-2g(x,2x)$$$$g(x,x^2+1)=\frac{e^{-x(x^2+1)}}{x+2(x^2+1)}=\frac{e^{-x^3-x}}{2x^2+x+2}$$$$g(x,2x)=\frac{e^{-x\left(2x\right)}}{x+2\left(2x\right)}=\frac{e^{-2x^2}}{5x}$$$$\frac{\partial g}{\partial x}(x,t)=\frac{-t{e}^{-xt}(x+2t)-e^{-xt}}{{(x+2t)}^2}=\frac{(-tx-2t^2)e^{-xt}}{{(x+2t)}^2}\Rightarrow$$$$F{}^{{}^{\prime }}\left(x\right)=-{\int }_{2x}^{x^2+1}\frac{(tx+2t^2)e^{-xt}}{{(x+2t)}^2}dt+2x\frac{e^{-x^3-x}}{2x^2+x+2}-2\frac{e^{-2x^2}}{5x}$$

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