What is the integral of 1/(12y-10y^2) ?

The integral of 1/(12y-10y^2) is 1/12 ln|y|-1/12 ln|5y-6|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(12y-10y^2)

\int \frac{1}{12 y - 10 y ^{2}} d y

\frac{1}{12} ln ∣ y ∣ - \frac{1}{12} ln ∣ 5 y - 6 ∣ + C

Solution steps

\int \frac{1}{12 y - 10 y ^{2}} d y

Take the partial fraction of

\frac{1}{12 y - 10 y ^{2}} : \frac{1}{12 y} - \frac{5}{12 ( 5 y - 6 )}

= \int \frac{1}{12 y} - \frac{5}{12 ( 5 y - 6 )} d y

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}{12 y} d y - \int \frac{5}{12 ( 5 y - 6 )} d y

\int \frac{1}{12 y} d y = \frac{1}{12} ln ∣ y ∣

\int \frac{5}{12 ( 5 y - 6 )} d y = \frac{1}{12} ln ∣ 5 y - 6 ∣

= \frac{1}{12} ln ∣ y ∣ - \frac{1}{12} ln ∣ 5 y - 6 ∣

Add a constant to the solution

= \frac{1}{12} ln ∣ y ∣ - \frac{1}{12} ln ∣ 5 y - 6 ∣ + C

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

f (x) = cos (cos (cos (x)))

n = 1 \sum \infty \frac{6}{( n + 1 ) !}

y^{^{''}} = a (y^{^{'}})^{2} + b

4 e^{3 x} cos (4 x)

What is the integral of 1/(12y-10y^2) ?
The integral of 1/(12y-10y^2) is 1/12 ln|y|-1/12 ln|5y-6|+C