What is the integral of 8/(x^2+6x+13) ?

The integral of 8/(x^2+6x+13) is 4arctan(1/2 (x+3))+C

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integral of 8/(x^2+6x+13)

\int \frac{8}{x ^{2} + 6 x + 13} d x

4 arctan (\frac{1}{2} (x + 3)) + C

Solution steps

\int \frac{8}{x ^{2} + 6 x + 13} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{x ^{2} + 6 x + 13} d x

Complete the square

x^{2} + 6 x + 13 : (x + 3)^{2} + 4

= 8 \cdot \int \frac{1}{( x + 3 ) ^{2} + 4} d x

Apply u-substitution

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

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

Apply Integral Substitution

= 8 \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

= 8 \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)

= 8 \cdot \frac{1}{2} arctan (v)

Substitute back

= 8 \cdot \frac{1}{2} arctan (\frac{x + 3}{2})

Simplify

8 \cdot \frac{1}{2} arctan (\frac{x + 3}{2}) : 4 arctan (\frac{1}{2} (x + 3))

= 4 arctan (\frac{1}{2} (x + 3))

Add a constant to the solution

= 4 arctan (\frac{1}{2} (x + 3)) + C

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What is the integral of 8/(x^2+6x+13) ?
The integral of 8/(x^2+6x+13) is 4arctan(1/2 (x+3))+C