What is the integral of 1/(y^2+3y+1) ?

The integral of 1/(y^2+3y+1) is -1/(sqrt(5))(ln|(2y+3)/(sqrt(5))+1|-ln|(2y+3)/(sqrt(5))-1|)+C

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integral of 1/(y^2+3y+1)

\int \frac{1}{y ^{2} + 3 y + 1} d y

- \frac{1}{5} (ln \frac{2 y + 3}{5} + 1 - ln \frac{2 y + 3}{5} - 1) + C

Solution steps

\int \frac{1}{y ^{2} + 3 y + 1} d y

Complete the square

y^{2} + 3 y + 1 : (y + \frac{3}{2})^{2} - \frac{5}{4}

= \int \frac{1}{( y + \frac{3}{2} ) ^{2} - \frac{5}{4}} d y

Apply u-substitution

: \int \frac{4}{4 u ^{2} - 5} d u

= \int \frac{4}{4 u ^{2} - 5} d u

Take the constant out:

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

= 4 \cdot \int \frac{1}{4 u ^{2} - 5} d u

Apply Integral Substitution

= 4 \cdot \int \frac{1}{2 5 ( v ^{2} - 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2 5} \cdot \int \frac{1}{v ^{2} - 1} d v

Factor

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

= 4 \cdot \frac{1}{2 5} \cdot \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2 5} (- \int \frac{1}{- v ^{2} + 1} d v)

Use the common integral:

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

= 4 \cdot \frac{1}{2 5} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= 4 \cdot \frac{1}{2 5} - \frac{ln \frac{2}{5} ( y + \frac{3}{2} ) + 1}{2} - \frac{ln \frac{2}{5} ( y + \frac{3}{2} ) - 1}{2}

Simplify

4 \cdot \frac{1}{2 5} - \frac{ln \frac{2}{5} ( y + \frac{3}{2} ) + 1}{2} - \frac{ln \frac{2}{5} ( y + \frac{3}{2} ) - 1}{2} : - \frac{1}{5} (ln \frac{2 y + 3}{5} + 1 - ln \frac{2 y + 3}{5} - 1)

= - \frac{1}{5} (ln \frac{2 y + 3}{5} + 1 - ln \frac{2 y + 3}{5} - 1)

Add a constant to the solution

= - \frac{1}{5} (ln \frac{2 y + 3}{5} + 1 - ln \frac{2 y + 3}{5} - 1) + C

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What is the integral of 1/(y^2+3y+1) ?
The integral of 1/(y^2+3y+1) is -1/(sqrt(5))(ln|(2y+3)/(sqrt(5))+1|-ln|(2y+3)/(sqrt(5))-1|)+C