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

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

Popular >

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

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

ln ∣ x ∣ - ln ∣ 2 x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1}{x - 2 x ^{2}} : \frac{1}{x} - \frac{2}{2 x - 1}

= \int \frac{1}{x} - \frac{2}{2 x - 1} d 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{1}{x} d x - \int \frac{2}{2 x - 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{2}{2 x - 1} d x = ln ∣ 2 x - 1 ∣

= ln ∣ x ∣ - ln ∣ 2 x - 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ 2 x - 1 ∣ + C

\int 2 x ln^{2} (x) d x

\int \frac{2 x}{5 + 3 x ^{2}} d x

\int \frac{8 + t + 6 t ^{2} - 12 t ^{3}}{( 3 t ^{2} + 4 ) ( t ^{2} + 7 )} d t

\int 5 e^{c o s (x)} sin (x) d x

\int x e^{x^{2} + 4} d x

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