What is the integral of ((8x-3))/((4x^2+(-3)x)) ?

The integral of ((8x-3))/((4x^2+(-3)x)) is ln|4x-3|+ln|x|+C

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integral of ((8x-3))/((4x^2+(-3)x))

\int \frac{( 8 x - 3 )}{( 4 x ^{2} + ( - 3 ) x )} d x

ln ∣ 4 x - 3 ∣ + ln ∣ x ∣ + C

Solution steps

\int \frac{8 x - 3}{4 x ^{2} + ( - 3 ) x} d x

Simplify

= \int \frac{8 x - 3}{4 x ^{2} - 3 x} d x

Expand

\frac{8 x - 3}{4 x ^{2} - 3 x} : \frac{8 x}{4 x ^{2} - 3 x} - \frac{3}{4 x ^{2} - 3 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{8 x}{4 x ^{2} - 3 x} d x - \int \frac{3}{4 x ^{2} - 3 x} d x

\int \frac{8 x}{4 x ^{2} - 3 x} d x = 2 ln ∣ 4 x - 3 ∣

\int \frac{3}{4 x ^{2} - 3 x} d x = - ln ∣ x ∣ + ln ∣ 4 x - 3 ∣

= 2 ln ∣ 4 x - 3 ∣ - (- ln ∣ x ∣ + ln ∣ 4 x - 3 ∣)

Simplify

= ln ∣ 4 x - 3 ∣ + ln ∣ x ∣

Add a constant to the solution

= ln ∣ 4 x - 3 ∣ + ln ∣ x ∣ + C

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

x \to 1 lim ((1 + \frac{8}{- 1})^{- x})

d x = x x^{2} - 16 d y

\frac{6}{5 x + 2}

\int 3 y d y

What is the integral of ((8x-3))/((4x^2+(-3)x)) ?
The integral of ((8x-3))/((4x^2+(-3)x)) is ln|4x-3|+ln|x|+C