What is the integral of 1/(3x^2+6x+6) ?

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

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integral of 1/(3x^2+6x+6)

\int \frac{1}{3 x ^{2} + 6 x + 6} d x

\frac{1}{3} arctan (x + 1) + C

Solution steps

\int \frac{1}{3 x ^{2} + 6 x + 6} d x

\frac{1}{3 x ^{2} + 6 x + 6} = \frac{1}{3} \frac{1}{x ^{2} + 2 x + 2}

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

Take the constant out:

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

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

Complete the square

x^{2} + 2 x + 2 : (x + 1)^{2} + 1

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

Apply u-substitution

: \int \frac{1}{u ^{2} + 1} d u

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

Use the common integral:

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

= \frac{1}{3} arctan (u)

Substitute back

u = x + 1

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

Add a constant to the solution

= \frac{1}{3} arctan (x + 1) + C

sin (x), x = 0, y = 0, x = \frac{π}{6}

y^{^{''}} + 4 y^{^{'}} + 4 y = x \cdot cos (2 x) + 3 e^{- 2 x}

\frac{d x}{d t} + 3 x = e^{- 2 t}

x \to 0 - lim ((1 + x)^{\frac{1}{x}})

f (x) = (4 x - x^{2})^{3}

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