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

The integral of (3x-1)/(4-ax) is 3/(a^2)(4-ax-4ln|4-ax|)+1/a ln|4-ax|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int \frac{3 x - 1}{4 - a x} d x

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

Solution steps

\int \frac{3 x - 1}{4 - a x} d x

Expand

\frac{3 x - 1}{4 - a x} : \frac{3 x}{4 - a x} - \frac{1}{4 - a 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{3 x}{4 - a x} d x - \int \frac{1}{4 - a x} d x

\int \frac{3 x}{4 - a x} d x = \frac{3}{a ^{2}} (4 - a x - 4 ln ∣ 4 - a x ∣)

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

= \frac{3}{a ^{2}} (4 - a x - 4 ln ∣ 4 - a x ∣) - (- \frac{1}{a} ln ∣ 4 - a x ∣)

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (sin (n x) e^{- n y})

y^{^{''}} + 11 y^{^{'}} + 18 y = 0

\int \frac{1}{16 x + 16 x} d x

y = x x, at x = 16

x \to 3 lim (10 - x)

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