What is the integral of 1/(2y-y^2) ?

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

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integral of 1/(2y-y^2)

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{1}{- u ^{2} + 1} d u

= \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

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

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

Substitute back

u = y - 1

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

Simplify

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

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

Add a constant to the solution

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

y = 2 x^{3} - 5 x + 1, (- 1, 4)

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h \to 0 lim (2)

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\int sec^{2} (5 x) tan^{4} (5 x) d x

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