What is the integral of 1/(3x^2-6x+18) ?

The integral of 1/(3x^2-6x+18) is 1/(3sqrt(5))arctan(1/(sqrt(5))(x-1))+C

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integral of 1/(3x^2-6x+18)

\int \frac{1}{3 x ^{2} - 6 x + 18} d x

\frac{1}{3 5} arctan (\frac{1}{5} (x - 1)) + C

Solution steps

\int \frac{1}{3 x ^{2} - 6 x + 18} d x

Complete the square

3 x^{2} - 6 x + 18 : 3 (x - 1)^{2} + 15

= \int \frac{1}{3 ( x - 1 ) ^{2} + 15} d x

Apply u-substitution

: \int \frac{1}{3 u ^{2} + 15} d u

= \int \frac{1}{3 u ^{2} + 15} d u

Apply Integral Substitution

= \int \frac{1}{3 5 ( 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}{3 5} \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}{3 5} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of 1/(3x^2-6x+18) ?
The integral of 1/(3x^2-6x+18) is 1/(3sqrt(5))arctan(1/(sqrt(5))(x-1))+C