What is the integral of ((5x+1))/((2x+1)(x-1)) ?

The integral of ((5x+1))/((2x+1)(x-1)) is 1/2 ln|2x+1|+2ln|x-1|+C

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integral of ((5x+1))/((2x+1)(x-1))

\int \frac{( 5 x + 1 )}{( 2 x + 1 ) ( x - 1 )} d x

\frac{1}{2} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{5 x + 1}{( 2 x + 1 ) ( x - 1 )} d x

Take the partial fraction of

\frac{5 x + 1}{( 2 x + 1 ) ( x - 1 )} : \frac{1}{2 x + 1} + \frac{2}{x - 1}

= \int \frac{1}{2 x + 1} + \frac{2}{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}{2 x + 1} d x + \int \frac{2}{x - 1} d x

\int \frac{1}{2 x + 1} d x = \frac{1}{2} ln ∣ 2 x + 1 ∣

\int \frac{2}{x - 1} d x = 2 ln ∣ x - 1 ∣

= \frac{1}{2} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x - 1 ∣ + C

y^{^{''''}} + 3 y^{^{''}} - 4 y = 0

arctan (\frac{x - 1}{x + 1})

\int 2 e^{2 x - 1} d x

\frac{6 s + 7}{s ^{2} - s - 6}

\int \frac{1}{( 2 x + 1 ) ^{5}} d x

What is the integral of ((5x+1))/((2x+1)(x-1)) ?
The integral of ((5x+1))/((2x+1)(x-1)) is 1/2 ln|2x+1|+2ln|x-1|+C