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

The integral of ((3x-6))/((3x+1)(x-1)) is 7/4 ln|3x+1|-3/4 ln|x-1|+C

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integral of ((3x-6))/((3x+1)(x-1))

integral

\frac{( 3 x - 6 )}{( 3 x + 1 ) ( x - 1 )}

\frac{7}{4} ln ∣ 3 x + 1 ∣ - \frac{3}{4} ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{3 x - 6}{( 3 x + 1 ) ( x - 1 )} d x

Take the partial fraction of

\frac{3 x - 6}{( 3 x + 1 ) ( x - 1 )} : \frac{21}{4 ( 3 x + 1 )} - \frac{3}{4 ( x - 1 )}

= \int \frac{21}{4 ( 3 x + 1 )} - \frac{3}{4 ( x - 1 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{21}{4 ( 3 x + 1 )} d x - \int \frac{3}{4 ( x - 1 )} d x

\int \frac{21}{4 ( 3 x + 1 )} d x = \frac{7}{4} ln ∣ 3 x + 1 ∣

\int \frac{3}{4 ( x - 1 )} d x = \frac{3}{4} ln ∣ x - 1 ∣

= \frac{7}{4} ln ∣ 3 x + 1 ∣ - \frac{3}{4} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{7}{4} ln ∣ 3 x + 1 ∣ - \frac{3}{4} ln ∣ x - 1 ∣ + C

\frac{x}{x ^{2} + y ^{2}}

e^{4} x - 3

1^{22} x \cdot x^{2} - 11

f (x) = x^{3} (3 x^{5} - 10)^{11}

\frac{x}{a ^{2}}

What is the integral of ((3x-6))/((3x+1)(x-1)) ?
The integral of ((3x-6))/((3x+1)(x-1)) is 7/4 ln|3x+1|-3/4 ln|x-1|+C