What is the integral of 1/((1-y)(1+y)) ?

The integral of 1/((1-y)(1+y)) is (ln|y+1|)/2-(ln|y-1|)/2+C

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integral of 1/((1-y)(1+y))

\int \frac{1}{( 1 - y ) ( 1 + y )} d y

\frac{ln ∣ y + 1 ∣}{2} - \frac{ln ∣ y - 1 ∣}{2} + C

Solution steps

\int \frac{1}{( 1 - y ) ( 1 + y )} d y

Expand

(1 - y) (1 + y) : 1 - y^{2}

= \int \frac{1}{1 - y ^{2}} d y

Use the common integral:

\int \frac{1}{1 - y ^{2}} d y = \frac{ln ∣ y + 1 ∣}{2} - \frac{ln ∣ y - 1 ∣}{2}

= \frac{ln ∣ y + 1 ∣}{2} - \frac{ln ∣ y - 1 ∣}{2}

Add a constant to the solution

= \frac{ln ∣ y + 1 ∣}{2} - \frac{ln ∣ y - 1 ∣}{2} + C

y^{^{''}} + 6 y^{^{'}} - 27 y = 0

\int \frac{x}{( 4 - 2 x ^{2} ) ^{3}} d x

\int \frac{x + 9}{x ^{2} + 4 x + 8} d x

\frac{d}{d x} (\frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8)

\int x^{2} x^{3} + 35 d x

What is the integral of 1/((1-y)(1+y)) ?
The integral of 1/((1-y)(1+y)) is (ln|y+1|)/2-(ln|y-1|)/2+C