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Question Number 67974 by mathmax by abdo last updated on 02/Sep/19

$$letF\left(x\right)={\int }_x^{x^2}\frac{arctan\left(xt\right)}{x^2+t^2}dtcalculateF{}^{{}^{\prime }}\left(x\right).$$

Commented by mathmax by abdo last updated on 03/Sep/19

$$wehaveg(x,t)=\frac{arctan\left(xt\right)}{x^2+t^2},u\left(x\right)=xandv\left(x\right)=x^2$$$$weusetheformulaeF{}^{{}^{\prime }}\left(x\right)={\int }_{u\left(x\right)}^{v\left(x\right)}\frac{\partial g}{\partial x}(x,t)dt+v^{{}^{\prime }}g(x,v)$$$$-u^{{}^{\prime }}g(x,u)={\int }_x^{x^2}\frac{\partial g}{\partial x}(x,t)dt+2x\times \frac{arctan\left(x^3\right)}{x^2+x^4}$$$$-\frac{arctan\left(x^2\right)}{2x^2}$$$$\frac{\partial g}{\partial x}(x,t)=\frac{\frac{t}{1+x^2{t}^2}(x^2+t^2)-2xarctan\left(xt\right)}{{(x^2+t^2)}^2}$$$$$$

Commented by mathmax by abdo last updated on 03/Sep/19

$$\Rightarrow F{}^{{}^{\prime }}\left(x\right)={\int }_x^{x^2}(\frac{t}{(1+x^2{t}^2)(x^2+t^2)}-\frac{2xarctan\left(xt\right)}{{(x^2+t^2)}^2})dt$$$$+\frac{2xarctan\left(x^3\right)}{x^2+x^4}-\frac{arctan\left(x^2\right)}{2x^2}....becontinued...$$

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