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

The integral of 1/(2y+y^2) is -1/2 ln|y+2|+1/2 ln|y|+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 + 2 ∣ + \frac{1}{2} ln ∣ y ∣ + 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

Factor

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \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 + 2 ∣ + \frac{1}{2} ln ∣ y ∣

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

Add a constant to the solution

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

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\int_{0}^{1} \frac{11}{4 y - 1} d y

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