What is the integral of 1/((5x-3)(3-4x)) ?

The integral of 1/((5x-3)(3-4x)) is 1/3 ln|5x-3|-1/3 ln|4x-3|+C

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integral of 1/((5x-3)(3-4x))

\int \frac{1}{( 5 x - 3 ) ( 3 - 4 x )} d x

\frac{1}{3} ln ∣ 5 x - 3 ∣ - \frac{1}{3} ln ∣ 4 x - 3 ∣ + C

Solution steps

\int \frac{1}{( 5 x - 3 ) ( 3 - 4 x )} d x

Take the partial fraction of

\frac{1}{( 5 x - 3 ) ( 3 - 4 x )} : \frac{5}{3 ( 5 x - 3 )} - \frac{4}{3 ( 4 x - 3 )}

= \int \frac{5}{3 ( 5 x - 3 )} - \frac{4}{3 ( 4 x - 3 )} 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{5}{3 ( 5 x - 3 )} d x - \int \frac{4}{3 ( 4 x - 3 )} d x

\int \frac{5}{3 ( 5 x - 3 )} d x = \frac{1}{3} ln ∣ 5 x - 3 ∣

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

= \frac{1}{3} ln ∣ 5 x - 3 ∣ - \frac{1}{3} ln ∣ 4 x - 3 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 5 x - 3 ∣ - \frac{1}{3} ln ∣ 4 x - 3 ∣ + C

\frac{\partial}{\partial x} (\frac{14 x}{7 x ^{2} + y ^{2} + 6})

y = 2 x, (- 3, 3)

\int_{0}^{\frac{π}{32}} cos (8 x) d x

ln (2 - x)

n = 0 \sum \infty 6 (- \frac{1}{3})^{n}

What is the integral of 1/((5x-3)(3-4x)) ?
The integral of 1/((5x-3)(3-4x)) is 1/3 ln|5x-3|-1/3 ln|4x-3|+C