What is the integral of 4/(3x-2) ?

The integral of 4/(3x-2) is 4/3 ln|3x-2|+C

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integral of 4/(3x-2)

integral

\frac{4}{3 x - 2}

\frac{4}{3} ln ∣ 3 x - 2 ∣ + C

Solution steps

\int \frac{4}{3 x - 2} 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}{3 x - 2} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= 4 \cdot \int \frac{1}{3 u} d u

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 \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 x - 2

= 4 \cdot \frac{1}{3} ln ∣ 3 x - 2 ∣

Simplify

4 \cdot \frac{1}{3} ln ∣ 3 x - 2 ∣ : \frac{4}{3} ln ∣ 3 x - 2 ∣

= \frac{4}{3} ln ∣ 3 x - 2 ∣

Add a constant to the solution

= \frac{4}{3} ln ∣ 3 x - 2 ∣ + C

\frac{1}{( 1 + cos ^{2} ( x ) )}

\frac{( y + 1 )}{y ^{3}}

\frac{tan ( x )}{2}

\frac{ln ( t )}{( 1 + t ^{2} )}

t^{- 2}