What is the integral of 1/(3x^2+6x+7) ?

The integral of 1/(3x^2+6x+7) is 1/(2sqrt(3))arctan((sqrt(3))/2 (x+1))+C

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integral of 1/(3x^2+6x+7)

\int \frac{1}{3 x ^{2} + 6 x + 7} d x

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

Solution steps

\int \frac{1}{3 x ^{2} + 6 x + 7} d x

Complete the square

3 x^{2} + 6 x + 7 : 3 (x + 1)^{2} + 4

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

Apply u-substitution

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

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

Apply Integral Substitution

= \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

= \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)

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

Substitute back

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

Add a constant to the solution

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

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What is the integral of 1/(3x^2+6x+7) ?
The integral of 1/(3x^2+6x+7) is 1/(2sqrt(3))arctan((sqrt(3))/2 (x+1))+C