What is the integral of 1/(x^2+3x+3) ?

The integral of 1/(x^2+3x+3) is 2/(sqrt(3))arctan(2/(sqrt(3))x+sqrt(3))+C

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integral of 1/(x^2+3x+3)

\int \frac{1}{x ^{2} + 3 x + 3} d x

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

= 4 \cdot \int \frac{1}{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

= 4 \cdot \frac{1}{2 3} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 4 \cdot \frac{1}{2 3} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

y^{''} (x) - 3 y^{'} (x) + 2 y = \frac{1}{1 + e ^{- x}}

x \to 0 lim (- arcsin (x))

\int_{0}^{1} 9 cos (\frac{π t}{2}) d t

\int \frac{1}{x ^{2} + 36} d x

What is the integral of 1/(x^2+3x+3) ?
The integral of 1/(x^2+3x+3) is 2/(sqrt(3))arctan(2/(sqrt(3))x+sqrt(3))+C