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

The integral of x/(x^2+3x+3) is 1/2 ln|4x^2+12x+12|-sqrt(3)arctan(2/(sqrt(3))x+sqrt(3))+C

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

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

\frac{1}{2} ln 4 x^{2} + 12 x + 12 - 3 arctan (\frac{2}{3} x + 3) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

\int \frac{2 u}{4 u ^{2} + 3} d u = \frac{1}{4} ln 4 u^{2} + 3

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

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

Substitute back

u = x + \frac{3}{2}

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

Simplify

2 (\frac{1}{4} ln 4 (x + \frac{3}{2})^{2} + 3 - \frac{3}{2} arctan (\frac{2}{3} (x + \frac{3}{2}))) : \frac{1}{2} ln 4 x^{2} + 12 x + 12 - 3 arctan (\frac{2}{3} x + 3)

= \frac{1}{2} ln 4 x^{2} + 12 x + 12 - 3 arctan (\frac{2}{3} x + 3)

Add a constant to the solution

= \frac{1}{2} ln 4 x^{2} + 12 x + 12 - 3 arctan (\frac{2}{3} x + 3) + C

8 tan^{2} (8 x)

\frac{d}{d x} (x^{2} - 2 x)

\int 3 y^{5} d y

\int_{- a}^{a} f (x) d x

\frac{d}{d x} (x^{5} + 37 x)

What is the integral of x/(x^2+3x+3) ?
The integral of x/(x^2+3x+3) is 1/2 ln|4x^2+12x+12|-sqrt(3)arctan(2/(sqrt(3))x+sqrt(3))+C