# d/dx of ʃ (superscript x)(subscript -x) (z+1)/(z+2)dz

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It is a lot easier if you use the fundamental theorem of Calculus

`(d)/(dx) int^(u(x))_(v(x)) f(t) dt = f(u(x))(du(x))/(dx) - f(v(u(x)))(dv(x))/(dx) `

`(d)/(dx) int^x_(-x) (z+1)/(z+2) dx = (x+1)/(x+2) (d(x))/(dx) - (-x+1)/(-x+2) (d(-x))/(dx) `

`= (x+1)/(x+2) - (-(x-1)/(x-2)) = ((x+1)(x-2)+(x-1)(x+2))/(x^2-4) = (x^2-x-2+x^2+x-2)/(x^2-4) `

`= (2x^2 - 4)/(x^2 - 4) = 2(x^2-2)/(x^2-4)`  Final answer

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You should test if the function is odd or even such that:

`f(-z) = (1-z)/(2+z)`

Notice that the function is not odd, nor even, hence, you need to evaluate the definite integral such that:

`int_(-x)^x (z+1)/(z+2) dz = int_(-x)^x (z+2-1)/(z+2) dz`

`int_(-x)^x (z+1)/(z+2) dz = int_(-x)^x (z+2)/(z+2) dz -int_(-x)^x (dz)/(z+2)`

`int_(-x)^x (z+1)/(z+2) dz = int_(-x)^x dz - int_(-x)^x (dz)/(z+2)`

`int_(-x)^x (z+1)/(z+2) dz = (z - ln|z+2|)_(-x)^x`

`int_(-x)^x (z+1)/(z+2) dz = x - ln|x+2| + x + ln|2-x|`

`int_(-x)^x (z+1)/(z+2) dz = 2x - ln|x+2| + ln|2-x|`

You need to differentiate the result with respect to x such that:

`(d(int_(-x)^x (z+1)/(z+2) dz))/(dx) = (d(2x - ln|x+2| + ln|2-x|))/(dx)`

`(d(int_(-x)^x (z+1)/(z+2) dz))/(dx) = 2 - 1/(x+2) - 1/(2-x)`

`(d(int_(-x)^x (z+1)/(z+2) dz))/(dx) = (-2(x^2 - 4) - 2 + x - x - 2)/(-(x^2 - 4))`

`(d(int_(-x)^x (z+1)/(z+2) dz))/(dx) = (-2x^2 + 8 - 4)/(-(x^2 - 4))`

`(d(int_(-x)^x (z+1)/(z+2) dz))/(dx) = -2(x^2 - 2)/(-(x^2 - 4))`

Hence, differentiating the definite integral with respect to x yields `(d(int_(-x)^x (z+1)/(z+2) dz))/(dx) = 2(x^2 - 2)/(x^2 - 4).`

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