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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\int (2 x)^{\frac{3}{2}} d x

\int t^{3} + t d t

\int y (1 - y) d y

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

\int \frac{y}{ln ( y )} d y

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