What is the integral of (3+16x)/(8x-4) ?

The integral of (3+16x)/(8x-4) is 3/8 ln|8x-4|+2x-1+ln(4|2x-1|)+C

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integral of (3+16x)/(8x-4)

\int \frac{3 + 16 x}{8 x - 4} d x

\frac{3}{8} ln ∣ 8 x - 4 ∣ + 2 x - 1 + ln (4 ∣ 2 x - 1 ∣) + C

Solution steps

\int \frac{3 + 16 x}{8 x - 4} d x

Expand

\frac{3 + 16 x}{8 x - 4} : \frac{3}{8 x - 4} + \frac{16 x}{8 x - 4}

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{3}{8 x - 4} d x + \int \frac{16 x}{8 x - 4} d x

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

\int \frac{16 x}{8 x - 4} d x = 2 x - 1 + ln (4 ∣ 2 x - 1 ∣)

= \frac{3}{8} ln ∣ 8 x - 4 ∣ + 2 x - 1 + ln (4 ∣ 2 x - 1 ∣)

Add a constant to the solution

= \frac{3}{8} ln ∣ 8 x - 4 ∣ + 2 x - 1 + ln (4 ∣ 2 x - 1 ∣) + C

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

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

(- \frac{1}{4 x ^{\frac{3}{2}}})^{^{'}}

\frac{d y}{d x} + y^{9} x + 9 y = 0

x \to 4 lim (x^{2} - 2 x - 8)

What is the integral of (3+16x)/(8x-4) ?
The integral of (3+16x)/(8x-4) is 3/8 ln|8x-4|+2x-1+ln(4|2x-1|)+C