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

let F(x)=∫_(2x) ^(x^2 +1) (e^(−xt) /(x+2t))dt calculate F^′ (x)

$${let}\:{F}\left({x}\right)=\int_{\mathrm{2}{x}} ^{{x}^{\mathrm{2}} +\mathrm{1}} \:\:\frac{{e}^{−{xt}} }{{x}+\mathrm{2}{t}}{dt}\:\:\:\:{calculate}\:{F}\:^{'} \left({x}\right) \\ $$

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

F(x)=∫_(2x) ^(x^2 +1) (e^(−xt) /(x+2t))dt we have u(x)=2x , v(x)=x^2 +1 ,g(x,t)=(e^(−xt) /(x+2t)) we aplly tbe formula ⇒ F^′ (x) =∫_(u(x)) ^(v(x)) (∂g/∂x)(x,t)dt +v^′ g(x,v) −u^′ g(x,u)=∫_(2x) ^(x^2 +1) (∂g/∂x)(x,t)dt +2xg(x,x^2 +1)−2g(x,2x) g(x,x^2 +1) =(e^(−x(x^2 +1)) /(x+2(x^2 +1))) =(e^(−x^3 −x) /(2x^2 +x +2)) g(x,2x) = (e^(−x(2x)) /(x+2(2x))) =(e^(−2x^2 ) /(5x)) (∂g/∂x)(x,t) =((−t e^(−xt) (x+2t)−e^(−xt) )/((x+2t)^2 )) =(((−tx−2t^2 )e^(−xt) )/((x+2t)^2 )) ⇒ F^′ (x) =−∫_(2x) ^(x^2 +1) (((tx+2t^2 )e^(−xt) )/((x+2t)^2 ))dt +2x (e^(−x^3 −x) /(2x^2 +x+2)) −2(e^(−2x^2 ) /(5x))

$${F}\left({x}\right)=\int_{\mathrm{2}{x}} ^{{x}^{\mathrm{2}} +\mathrm{1}} \:\frac{{e}^{−{xt}} }{{x}+\mathrm{2}{t}}{dt}\:\:{we}\:{have}\:{u}\left({x}\right)=\mathrm{2}{x}\:,\:{v}\left({x}\right)={x}^{\mathrm{2}} \:+\mathrm{1}\:,{g}\left({x},{t}\right)=\frac{{e}^{−{xt}} }{{x}+\mathrm{2}{t}} \\ $$$${we}\:{aplly}\:{tbe}\:{formula}\:\Rightarrow\:{F}\:^{'} \left({x}\right)\:=\int_{{u}\left({x}\right)} ^{{v}\left({x}\right)} \:\frac{\partial{g}}{\partial{x}}\left({x},{t}\right){dt}\:+{v}^{'} {g}\left({x},{v}\right) \\ $$$$−{u}^{'} {g}\left({x},{u}\right)=\int_{\mathrm{2}{x}} ^{{x}^{\mathrm{2}} \:+\mathrm{1}} \:\frac{\partial{g}}{\partial{x}}\left({x},{t}\right){dt}\:+\mathrm{2}{xg}\left({x},{x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{2}{g}\left({x},\mathrm{2}{x}\right) \\ $$$${g}\left({x},{x}^{\mathrm{2}} +\mathrm{1}\right)\:=\frac{{e}^{−{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)} }{{x}+\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:=\frac{{e}^{−{x}^{\mathrm{3}} −{x}} }{\mathrm{2}{x}^{\mathrm{2}} \:+{x}\:+\mathrm{2}} \\ $$$${g}\left({x},\mathrm{2}{x}\right)\:=\:\frac{{e}^{−{x}\left(\mathrm{2}{x}\right)} }{{x}+\mathrm{2}\left(\mathrm{2}{x}\right)}\:=\frac{{e}^{−\mathrm{2}{x}^{\mathrm{2}} } }{\mathrm{5}{x}} \\ $$$$\frac{\partial{g}}{\partial{x}}\left({x},{t}\right)\:=\frac{−{t}\:{e}^{−{xt}} \left({x}+\mathrm{2}{t}\right)−{e}^{−{xt}} }{\left({x}+\mathrm{2}{t}\right)^{\mathrm{2}} }\:=\frac{\left(−{tx}−\mathrm{2}{t}^{\mathrm{2}} \right){e}^{−{xt}} }{\left({x}+\mathrm{2}{t}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${F}\:^{'} \left({x}\right)\:=−\int_{\mathrm{2}{x}} ^{{x}^{\mathrm{2}} \:+\mathrm{1}} \:\:\frac{\left({tx}+\mathrm{2}{t}^{\mathrm{2}} \right){e}^{−{xt}} }{\left({x}+\mathrm{2}{t}\right)^{\mathrm{2}} }{dt}\:+\mathrm{2}{x}\:\frac{{e}^{−{x}^{\mathrm{3}} −{x}} }{\mathrm{2}{x}^{\mathrm{2}} \:+{x}+\mathrm{2}}\:−\mathrm{2}\frac{{e}^{−\mathrm{2}{x}^{\mathrm{2}} } }{\mathrm{5}{x}} \\ $$

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