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

The integral of 2/(sqrt(3x^2-6x+18)) is 2/(sqrt(3))ln|(x-1}{sqrt(5)}+sqrt(\frac{x^2-2x+6)/5)|+C

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

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

\frac{2}{3} ln \frac{x - 1}{5} + \frac{x ^{2} - 2 x + 6}{5} + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{sec ( v )}{3} d v

= 2 \cdot \int \frac{sec ( v )}{3} d v

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int sec (v) d v

Use the common integral:

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= 2 \cdot \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 2 \cdot \frac{1}{3} ln tan (arctan (\frac{3}{15} (x - 1))) + sec (arctan (\frac{3}{15} (x - 1)))

Simplify

2 \cdot \frac{1}{3} ln tan (arctan (\frac{3}{15} (x - 1))) + sec (arctan (\frac{3}{15} (x - 1))) : \frac{2}{3} ln \frac{x - 1}{5} + \frac{x ^{2} - 2 x + 6}{5}

= \frac{2}{3} ln \frac{x - 1}{5} + \frac{x ^{2} - 2 x + 6}{5}

Add a constant to the solution

= \frac{2}{3} ln \frac{x - 1}{5} + \frac{x ^{2} - 2 x + 6}{5} + C

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What is the integral of 2/(sqrt(3x^2-6x+18)) ?
The integral of 2/(sqrt(3x^2-6x+18)) is 2/(sqrt(3))ln|(x-1}{sqrt(5)}+sqrt(\frac{x^2-2x+6)/5)|+C