What is the integral of 4/(sqrt(9x^2-6x)) ?

The integral of 4/(sqrt(9x^2-6x)) is 4/3 ln|sqrt(9x^2-6x)+3x-1|+C

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integral of 4/(sqrt(9x^2-6x))

\int \frac{4}{9 x ^{2} - 6 x} d x

\frac{4}{3} ln 9 x^{2} - 6 x + 3 x - 1 + C

Solution steps

\int \frac{4}{9 x ^{2} - 6 x} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{9 x ^{2} - 6 x} d x

Complete the square

9 x^{2} - 6 x : 9 (x - \frac{1}{3})^{2} - 1

= 4 \cdot \int \frac{1}{9 ( x - \frac{1}{3} ) ^{2} - 1} d x

Apply u-substitution

: \int \frac{1}{9 u ^{2} - 1} d u

= 4 \cdot \int \frac{1}{9 u ^{2} - 1} d u

Apply Trigonometric Substitution

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

= 4 \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

= 4 \cdot \frac{1}{3} \cdot \int sec (v) d v

Use the common integral:

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

= 4 \cdot \frac{1}{3} ln ∣ tan (v) + sec (v) ∣

Substitute back

= 4 \cdot \frac{1}{3} ln tan (arcsec (3 (x - \frac{1}{3}))) + sec (arcsec (3 (x - \frac{1}{3})))

Simplify

4 \cdot \frac{1}{3} ln tan (arcsec (3 (x - \frac{1}{3}))) + sec (arcsec (3 (x - \frac{1}{3}))) : \frac{4}{3} ln 9 x^{2} - 6 x + 3 x - 1

= \frac{4}{3} ln 9 x^{2} - 6 x + 3 x - 1

Add a constant to the solution

= \frac{4}{3} ln 9 x^{2} - 6 x + 3 x - 1 + C

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What is the integral of 4/(sqrt(9x^2-6x)) ?
The integral of 4/(sqrt(9x^2-6x)) is 4/3 ln|sqrt(9x^2-6x)+3x-1|+C