What is the integral of (19x^2+50x-25)/(3x^3-5x^2) ?

The integral of (19x^2+50x-25)/(3x^3-5x^2) is -7ln|x|-5/x+40/3 ln|3x-5|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (19x^2+50x-25)/(3x^3-5x^2)

\int \frac{19 x ^{2} + 50 x - 25}{3 x ^{3} - 5 x ^{2}} d x

- 7 ln ∣ x ∣ - \frac{5}{x} + \frac{40}{3} ln ∣ 3 x - 5 ∣ + C

Solution steps

\int \frac{19 x ^{2} + 50 x - 25}{3 x ^{3} - 5 x ^{2}} d x

Take the partial fraction of

\frac{19 x ^{2} + 50 x - 25}{3 x ^{3} - 5 x ^{2}} : - \frac{7}{x} + \frac{5}{x ^{2}} + \frac{40}{3 x - 5}

= \int - \frac{7}{x} + \frac{5}{x ^{2}} + \frac{40}{3 x - 5} 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{7}{x} d x + \int \frac{5}{x ^{2}} d x + \int \frac{40}{3 x - 5} d x

\int \frac{7}{x} d x = 7 ln ∣ x ∣

\int \frac{5}{x ^{2}} d x = - \frac{5}{x}

\int \frac{40}{3 x - 5} d x = \frac{40}{3} ln ∣ 3 x - 5 ∣

= - 7 ln ∣ x ∣ - \frac{5}{x} + \frac{40}{3} ln ∣ 3 x - 5 ∣

Add a constant to the solution

= - 7 ln ∣ x ∣ - \frac{5}{x} + \frac{40}{3} ln ∣ 3 x - 5 ∣ + C

\int (4 \frac{sin ( 2 x )}{sin ( x )}) d x

\int 9 x^{3} + 7 3 x^{2} d x

\int \frac{x ^{4} + x ^{3} + x + 5}{x + 1} d x

\int \frac{1}{2 + sin ( θ ) + cos ( θ )} d θ

\int 4 sinh (x) + x^{7} d x

What is the integral of (19x^2+50x-25)/(3x^3-5x^2) ?
The integral of (19x^2+50x-25)/(3x^3-5x^2) is -7ln|x|-5/x+40/3 ln|3x-5|+C