What is the integral of (6x)/(x^2+4x+13) ?

The integral of (6x)/(x^2+4x+13) is 3ln| 1/9 (x+2)^2+1|-4arctan(1/3 (x+2))+C

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integral of (6x)/(x^2+4x+13)

\int \frac{6 x}{x ^{2} + 4 x + 13} d x

3 ln \frac{1}{9} (x + 2)^{2} + 1 - 4 arctan (\frac{1}{3} (x + 2)) + C

Solution steps

\int \frac{6 x}{x ^{2} + 4 x + 13} d x

Take the constant out:

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

= 6 \cdot \int \frac{x}{x ^{2} + 4 x + 13} d x

Complete the square

x^{2} + 4 x + 13 : (x + 2)^{2} + 9

= 6 \cdot \int \frac{x}{( x + 2 ) ^{2} + 9} d x

Apply u-substitution

= 6 \cdot \int \frac{u - 2}{u ^{2} + 9} d u

Apply Integral Substitution

= 6 \cdot \int \frac{3 v - 2}{3 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 6 \cdot \frac{1}{3} \cdot \int \frac{3 v - 2}{v ^{2} + 1} d v

Expand

\frac{3 v - 2}{v ^{2} + 1} : \frac{3 v}{v ^{2} + 1} - \frac{2}{v ^{2} + 1}

Apply the Sum Rule:

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

= 6 \cdot \frac{1}{3} (\int \frac{3 v}{v ^{2} + 1} d v - \int \frac{2}{v ^{2} + 1} d v)

\int \frac{3 v}{v ^{2} + 1} d v = \frac{3}{2} ln v^{2} + 1

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

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

Substitute back

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

Simplify

6 \cdot \frac{1}{3} (\frac{3}{2} ln (\frac{x + 2}{3})^{2} + 1 - 2 arctan (\frac{x + 2}{3})) : 3 ln \frac{1}{9} (x + 2)^{2} + 1 - 4 arctan (\frac{1}{3} (x + 2))

= 3 ln \frac{1}{9} (x + 2)^{2} + 1 - 4 arctan (\frac{1}{3} (x + 2))

Add a constant to the solution

= 3 ln \frac{1}{9} (x + 2)^{2} + 1 - 4 arctan (\frac{1}{3} (x + 2)) + C

What is the integral of (6x)/(x^2+4x+13) ?
The integral of (6x)/(x^2+4x+13) is 3ln| 1/9 (x+2)^2+1|-4arctan(1/3 (x+2))+C