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

The integral of (2x)/(x^2+6x+13) is ln|x^2+6x+13|-3arctan(1/2 (x+3))+C

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

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

ln x^{2} + 6 x + 13 - 3 arctan (\frac{1}{2} (x + 3)) + C

Solution steps

\int \frac{2 x}{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

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

Complete the square

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

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

Apply u-substitution

: \int \frac{u - 3}{u ^{2} + 4} d u

= 2 \cdot \int \frac{u - 3}{u ^{2} + 4} d u

Expand

\frac{u - 3}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} - \frac{3}{u ^{2} + 4}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int \frac{u}{u ^{2} + 4} d u - \int \frac{3}{u ^{2} + 4} d u)

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

\int \frac{3}{u ^{2} + 4} d u = \frac{3}{2} arctan (\frac{u}{2})

= 2 (\frac{1}{2} ln u^{2} + 4 - \frac{3}{2} arctan (\frac{u}{2}))

Substitute back

u = x + 3

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

Simplify

2 (\frac{1}{2} ln (x + 3)^{2} + 4 - \frac{3}{2} arctan (\frac{x + 3}{2})) : ln x^{2} + 6 x + 13 - 3 arctan (\frac{1}{2} (x + 3))

= ln x^{2} + 6 x + 13 - 3 arctan (\frac{1}{2} (x + 3))

Add a constant to the solution

= ln x^{2} + 6 x + 13 - 3 arctan (\frac{1}{2} (x + 3)) + C

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