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

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

Popular >

integral of (2x+1)/(3x^2+3x-2)

\int \frac{2 x + 1}{3 x ^{2} + 3 x - 2} d x

\frac{1}{3} ln 3 x^{2} + 3 x - 2 + C

Solution steps

\int \frac{2 x + 1}{3 x ^{2} + 3 x - 2} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 x^{2} + 3 x - 2

= \frac{1}{3} ln 3 x^{2} + 3 x - 2

Add a constant to the solution

= \frac{1}{3} ln 3 x^{2} + 3 x - 2 + C

\frac{\partial}{\partial x} (sin (\frac{4 x}{y}))

\int x^{2} - 17 d x

\int cos^{8} (3 x) sin^{3} (3 x) d x

(3 x - 3 x^{2}) tan (x)

9 (t + 1) (\frac{d y}{d t}) - 6 y = 18 t

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