let-F-x-2x-x-2-1-e-xt-x-2t-dt-calculate-F-x-

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

l e t F (x) = \int_{2 x}^{x^{2} + 1} \frac{e^{- x t}}{x + 2 t} d t c a l c u l a t e F^{^{'}} (x)

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

F (x) = \int_{2 x}^{x^{2} + 1} \frac{e^{- x t}}{x + 2 t} d t w e h a v e u (x) = 2 x, v (x) = x^{2} + 1, g (x, t) = \frac{e^{- x t}}{x + 2 t}

w e a p l l y t b e f o r m u l a \Rightarrow F^{^{'}} (x) = \int_{u (x)}^{v (x)} \frac{\partial g}{\partial x} (x, t) d t + v^{^{'}} g (x, v)

- u^{^{'}} g (x, u) = \int_{2 x}^{x^{2} + 1} \frac{\partial g}{\partial x} (x, t) d t + 2 x g (x, x^{2} + 1) - 2 g (x, 2 x)

g (x, x^{2} + 1) = \frac{e^{- x (x^{2} + 1)}}{x + 2 (x^{2} + 1)} = \frac{e^{- x^{3} - x}}{2 x^{2} + x + 2}

g (x, 2 x) = \frac{e^{- x (2 x)}}{x + 2 (2 x)} = \frac{e^{- 2 x^{2}}}{5 x}

\frac{\partial g}{\partial x} (x, t) = \frac{- t e^{- x t} (x + 2 t) - e^{- x t}}{{(x + 2 t)}^{2}} = \frac{(- t x - 2 t^{2}) e^{- x t}}{{(x + 2 t)}^{2}} \Rightarrow

F^{^{'}} (x) = - \int_{2 x}^{x^{2} + 1} \frac{(t x + 2 t^{2}) e^{- x t}}{{(x + 2 t)}^{2}} d t + 2 x \frac{e^{- x^{3} - x}}{2 x^{2} + x + 2} - 2 \frac{e^{- 2 x^{2}}}{5 x}