How do you integrate $\frac{1}{y (1 - y)}$ $1/(y(1-y))$ ?

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How do you integrate $\frac{1}{y (1 - y)}$ $1/(y(1-y))$ ?

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Answer 1 · 2015-05-18T19:43:32

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Answer 2 · 2015-05-18T19:53:24

$\int \frac{1}{y (1 - y)} d y = - \int \frac{1}{y^{2} - y + {(- \frac{1}{2})}^{2} - {(- \frac{1}{2})}^{2}} d y$ $int1/(y(1-y))dy=-int1/(y^2-y+(-1/2)^2-(-1/2)^2)dy$

$\Rightarrow - \int \frac{1}{{(y - \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} d y = - \frac{1}{{(\frac{1}{2})}^{2}} \int \frac{1}{{(\frac{y - \frac{1}{2}}{\frac{1}{2}})}^{2} - 1} d y = - 4 \int \frac{1}{{(2 y - 1)}^{2} - 1} d y$ $=>-int1/((y-1/2)^2-(1/2)^2)dy=-1/(1/2)^2int1/(((y-1/2)/(1/2))^2-1)dy=-4int1/((2y-1)^2-1)dy$

$\Rightarrow \frac{1}{2} \cdot - 4 a r tanh (2 y - 1) + C = - ln ∣ ∣ ∣ \frac{1 - 2 y + 1}{1 + 2 y - 1} ∣ ∣ ∣ + C = ln ∣ ∣ ∣ \frac{y}{1 - y} ∣ ∣ ∣ + C$ $=>1/2*-4artanh(2y-1) + C = -ln|(1-2y+1)/(1+2y-1)| + C=ln|(y)/(1-y)| + C$

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How do you integrate $\frac{1}{y (1 - y)}$ $1/(y(1-y))$ ?

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How do you integrate 1y(1−y)1/(y(1-y))?

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How do you integrate $\frac{1}{y (1 - y)}$ $1/(y(1-y))$ ?