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

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on

Jul 24, 2020, 6:28:07 PM

Created by

on

Dec 27, 2013, 8:10:29 PM

\int_{a}^{b} \frac{1}{u z^{2} + v z + w} d z = (\frac{2}{\sqrt{4 u w - v^{2}}}) \cdot {tan}^{- 1} (\frac{2 u b + v}{\sqrt{4 u w - v^{2}}}) - (\frac{2}{\sqrt{4 u w - v^{2}}}) \cdot {tan}^{- 1} (\frac{2 u a + v}{\sqrt{4 u w - v^{2}}})

$int_a^b1/(uz^2 + vz+ w)dz = (2/sqrt(4 "u" "w" - "v" ^2))*tan^-1((2 "u" "b" + "v" )/sqrt(4 "u" "w" - "v" ^2)) - (2/sqrt(4 "u" "w" - "v" ^2))*tan^-1((2 "u" "a" + "v" )/sqrt(4 "u" "w" - "v" ^2))$

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

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∫ba1uz2+vz+wdz∫ba1uz2+vz+wdzint_a^b1/(uz^2 + vz+ w)dz

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