What is the integral of (17x+3)/((4x+1)(x-1)) ?

The integral of (17x+3)/((4x+1)(x-1)) is 1/4 ln|4x+1|+4ln|x-1|+C

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integral of (17x+3)/((4x+1)(x-1))

\int \frac{17 x + 3}{( 4 x + 1 ) ( x - 1 )} d x

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

Solution steps

\int \frac{17 x + 3}{( 4 x + 1 ) ( x - 1 )} d x

Take the partial fraction of

\frac{17 x + 3}{( 4 x + 1 ) ( x - 1 )} : \frac{1}{4 x + 1} + \frac{4}{x - 1}

= \int \frac{1}{4 x + 1} + \frac{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{1}{4 x + 1} d x + \int \frac{4}{x - 1} d x

\int \frac{1}{4 x + 1} d x = \frac{1}{4} ln ∣ 4 x + 1 ∣

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

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

Add a constant to the solution

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

\int 4 + 16 x^{2} d x

\int_{1}^{9} y^{\frac{1}{2}} ln (y) d y

\int \frac{v}{( v - 1 ) ^{2}} d v

\int e^{x + 3} d x

t^{2} \cdot \frac{d y}{d t} + y = 0

What is the integral of (17x+3)/((4x+1)(x-1)) ?
The integral of (17x+3)/((4x+1)(x-1)) is 1/4 ln|4x+1|+4ln|x-1|+C