What is the integral of (2+2x)/(x-4x(1+x)) ?

The integral of (2+2x)/(x-4x(1+x)) is 1/6 ln|4x+3|-2/3 ln(4|x|)+C

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integral of (2+2x)/(x-4x(1+x))

\int \frac{2 + 2 x}{x - 4 x ( 1 + x )} d x

\frac{1}{6} ln ∣ 4 x + 3 ∣ - \frac{2}{3} ln (4 ∣ x ∣) + C

Solution steps

\int \frac{2 + 2 x}{x - 4 x ( 1 + x )} d x

Expand

\frac{2 + 2 x}{x - 4 x ( 1 + x )} : \frac{2}{x - 4 x ( 1 + x )} + \frac{2 x}{x - 4 x ( 1 + 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{2}{x - 4 x ( 1 + x )} d x + \int \frac{2 x}{x - 4 x ( 1 + x )} d x

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

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

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

Simplify

2 (\frac{1}{3} ln ∣ 4 x + 3 ∣ - \frac{1}{3} ln (4 ∣ x ∣)) - \frac{1}{2} ln ∣ 4 x + 3 ∣ : \frac{1}{6} ln ∣ 4 x + 3 ∣ - \frac{2}{3} ln (4 ∣ x ∣)

= \frac{1}{6} ln ∣ 4 x + 3 ∣ - \frac{2}{3} ln (4 ∣ x ∣)

Add a constant to the solution

= \frac{1}{6} ln ∣ 4 x + 3 ∣ - \frac{2}{3} ln (4 ∣ x ∣) + C

What is the integral of (2+2x)/(x-4x(1+x)) ?
The integral of (2+2x)/(x-4x(1+x)) is 1/6 ln|4x+3|-2/3 ln(4|x|)+C