What is the integral of 1/(y^2+4y+8) ?

The integral of 1/(y^2+4y+8) is 1/2 arctan((y+2)/2)+C

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integral of 1/(y^2+4y+8)

\int \frac{1}{y ^{2} + 4 y + 8} d y

\frac{1}{2} arctan (\frac{y + 2}{2}) + C

Solution steps

\int \frac{1}{y ^{2} + 4 y + 8} d y

Complete the square

y^{2} + 4 y + 8 : (y + 2)^{2} + 4

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 4} d u

= \int \frac{1}{u ^{2} + 4} d u

Apply Integral Substitution

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

= \frac{1}{2} arctan (v)

Substitute back

= \frac{1}{2} arctan (\frac{y + 2}{2})

Add a constant to the solution

= \frac{1}{2} arctan (\frac{y + 2}{2}) + C

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What is the integral of 1/(y^2+4y+8) ?
The integral of 1/(y^2+4y+8) is 1/2 arctan((y+2)/2)+C