What is the integral of (x+4)/(9x^2-6x+1) ?

The integral of (x+4)/(9x^2-6x+1) is 1/9 ln|3x-1|-(13)/(9(3x-1))+C

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integral of (x+4)/(9x^2-6x+1)

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

\frac{1}{9} ln ∣ 3 x - 1 ∣ - \frac{13}{9 ( 3 x - 1 )} + C

Solution steps

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

Take the partial fraction of

\frac{x + 4}{9 x ^{2} - 6 x + 1} : \frac{1}{3 ( 3 x - 1 )} + \frac{13}{3 ( 3 x - 1 ) ^{2}}

= \int \frac{1}{3 ( 3 x - 1 )} + \frac{13}{3 ( 3 x - 1 ) ^{2}} 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{1}{3 ( 3 x - 1 )} d x + \int \frac{13}{3 ( 3 x - 1 ) ^{2}} d x

\int \frac{1}{3 ( 3 x - 1 )} d x = \frac{1}{9} ln ∣ 3 x - 1 ∣

\int \frac{13}{3 ( 3 x - 1 ) ^{2}} d x = - \frac{13}{9 ( 3 x - 1 )}

= \frac{1}{9} ln ∣ 3 x - 1 ∣ - \frac{13}{9 ( 3 x - 1 )}

Add a constant to the solution

= \frac{1}{9} ln ∣ 3 x - 1 ∣ - \frac{13}{9 ( 3 x - 1 )} + C

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What is the integral of (x+4)/(9x^2-6x+1) ?
The integral of (x+4)/(9x^2-6x+1) is 1/9 ln|3x-1|-(13)/(9(3x-1))+C