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

The integral of 1/((3x+4)(4x-1)) is -1/19 ln|3x+4|+1/19 ln|12x-3|+C

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

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

- \frac{1}{19} ln ∣ 3 x + 4 ∣ + \frac{1}{19} ln ∣ 12 x - 3 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( 4 u - 19 )} d u

= \int \frac{1}{u ( 4 u - 19 )} d u

Take the partial fraction of

\frac{1}{u ( 4 u - 19 )} : - \frac{1}{19 u} + \frac{4}{19 ( 4 u - 19 )}

= \int - \frac{1}{19 u} + \frac{4}{19 ( 4 u - 19 )} d u

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}{19 u} d u + \int \frac{4}{19 ( 4 u - 19 )} d u

\int \frac{1}{19 u} d u = \frac{1}{19} ln ∣ u ∣

\int \frac{4}{19 ( 4 u - 19 )} d u = \frac{1}{19} ln ∣ 4 u - 19 ∣

= - \frac{1}{19} ln ∣ u ∣ + \frac{1}{19} ln ∣ 4 u - 19 ∣

Substitute back

u = 3 x + 4

= - \frac{1}{19} ln ∣ 3 x + 4 ∣ + \frac{1}{19} ln ∣ 4 (3 x + 4) - 19 ∣

Simplify

- \frac{1}{19} ln ∣ 3 x + 4 ∣ + \frac{1}{19} ln ∣ 4 (3 x + 4) - 19 ∣ : - \frac{1}{19} ln ∣ 3 x + 4 ∣ + \frac{1}{19} ln ∣ 12 x - 3 ∣

= - \frac{1}{19} ln ∣ 3 x + 4 ∣ + \frac{1}{19} ln ∣ 12 x - 3 ∣

Add a constant to the solution

= - \frac{1}{19} ln ∣ 3 x + 4 ∣ + \frac{1}{19} ln ∣ 12 x - 3 ∣ + C

\frac{d}{d x} (4 x^{5} - 3 x^{4} + 2 x^{3} - x + 100)

(1 - x) d y - y^{2} d x = 0

\int \frac{32 x ^{2}}{( x - 24 ) ( x + 8 ) ^{2}} d x

\frac{d x}{d t} = \frac{1}{x e ^{t + 5 x}}

\int \frac{x ^{2}}{x ^{3} - 1} d x

What is the integral of 1/((3x+4)(4x-1)) ?
The integral of 1/((3x+4)(4x-1)) is -1/19 ln|3x+4|+1/19 ln|12x-3|+C