What is the integral of ((37x+5))/((6x+1)(x-1)) ?

The integral of ((37x+5))/((6x+1)(x-1)) is 1/6 ln|6x+1|+6ln|x-1|+C

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integral of ((37x+5))/((6x+1)(x-1))

\int \frac{( 37 x + 5 )}{( 6 x + 1 ) ( x - 1 )} d x

\frac{1}{6} ln ∣ 6 x + 1 ∣ + 6 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{37 x + 5}{( 6 x + 1 ) ( x - 1 )} d x

Take the partial fraction of

\frac{37 x + 5}{( 6 x + 1 ) ( x - 1 )} : \frac{1}{6 x + 1} + \frac{6}{x - 1}

= \int \frac{1}{6 x + 1} + \frac{6}{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{1}{6 x + 1} d x + \int \frac{6}{x - 1} d x

\int \frac{1}{6 x + 1} d x = \frac{1}{6} ln ∣ 6 x + 1 ∣

\int \frac{6}{x - 1} d x = 6 ln ∣ x - 1 ∣

= \frac{1}{6} ln ∣ 6 x + 1 ∣ + 6 ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{1}{6} ln ∣ 6 x + 1 ∣ + 6 ln ∣ x - 1 ∣ + C

\int (1 + 4 x^{2})^{\frac{1}{2}} d x

2 (ln (y)) y^{^{'}} - x^{3} y = 0

\int (- 3) d x

\int e^{s i n (θ)} d θ

f (x) = 4 + 3 x - 2 x^{2}, at x = 2

What is the integral of ((37x+5))/((6x+1)(x-1)) ?
The integral of ((37x+5))/((6x+1)(x-1)) is 1/6 ln|6x+1|+6ln|x-1|+C