$\int_{a}^{b} \frac{1}{1 + z^{2}} d z$ $int_a^b1/(1+z^2)dz$

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Jul 24, 2020, 6:28:07 PM

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Dec 27, 2013, 8:10:29 PM

\int_{a}^{b} \frac{1}{1 + z^{2}} d z = {tan}^{- 1} (b) - {tan}^{- 1} (a) + 0.0

$int_a^b1/(1+z^2)dz = tan^-1( "b" ) - tan^-1( "a" ) + 0.0$

a

b

C

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This equation calculates the definitie integral for the expression F(z) = 1/(1+z^2). It returns an answer expressed in units of radians in the range of -pi/2 to pi/2.

This is of the form containing the binomial 1+/- x^n

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$\int_{a}^{b} \frac{1}{1 + z^{2}} d z$ $int_a^b1/(1+z^2)dz$

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∫ba11+z2dz∫ba11+z2dzint_a^b1/(1+z^2)dz

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$\int_{a}^{b} \frac{1}{1 + z^{2}} d z$ $int_a^b1/(1+z^2)dz$