What is the integral of 1/(3x^2+6x+78) ?

The integral of 1/(3x^2+6x+78) is 1/15 arctan((x+1)/5)+C

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integral of 1/(3x^2+6x+78)

\int \frac{1}{3 x ^{2} + 6 x + 78} d x

\frac{1}{15} arctan (\frac{x + 1}{5}) + C

Solution steps

\int \frac{1}{3 x ^{2} + 6 x + 78} d x

Complete the square

3 x^{2} + 6 x + 78 : 3 (x + 1)^{2} + 75

= \int \frac{1}{3 ( x + 1 ) ^{2} + 75} d x

Apply u-substitution

= \int \frac{1}{3 u ^{2} + 75} d u

Apply Integral Substitution

= \int \frac{1}{15 ( 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}{15} \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}{15} arctan (v)

Substitute back

= \frac{1}{15} arctan (\frac{3}{5 3} (x + 1))

Simplify

\frac{1}{15} arctan (\frac{3}{5 3} (x + 1)) : \frac{1}{15} arctan (\frac{x + 1}{5})

= \frac{1}{15} arctan (\frac{x + 1}{5})

Add a constant to the solution

= \frac{1}{15} arctan (\frac{x + 1}{5}) + C

What is the integral of 1/(3x^2+6x+78) ?
The integral of 1/(3x^2+6x+78) is 1/15 arctan((x+1)/5)+C