What is the integral of (3x)/((x^2+4x+6)) ?

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

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integral of (3x)/((x^2+4x+6))

integral

\frac{3 x}{( x ^{2} + 4 x + 6 )}

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

Solution steps

\int \frac{3 x}{x ^{2} + 4 x + 6} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{x ^{2} + 4 x + 6} d x

Complete the square

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

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

Apply u-substitution

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

Apply Integral Substitution

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

Expand

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

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

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

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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