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

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

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integral of 4/((x^2+4x+8))

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

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

Solution steps

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

Take the constant out:

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

= 4 \cdot \int \frac{1}{x ^{2} + 4 x + 8} d x

Complete the square

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

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

Apply u-substitution

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

Apply Integral Substitution

= 4 \cdot \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

= 4 \cdot \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)

= 4 \cdot \frac{1}{2} arctan (v)

Substitute back

= 4 \cdot \frac{1}{2} arctan (\frac{x + 2}{2})

Simplify

4 \cdot \frac{1}{2} arctan (\frac{x + 2}{2}) : 2 arctan (\frac{1}{2} (x + 2))

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

Add a constant to the solution

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

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