What is the integral of (6x^4+x-3)/(2x^3+2x^2-12x) ?

The integral of (6x^4+x-3)/(2x^3+2x^2-12x) is (3x^2)/2-3x+1/4 ln|x|+19/4 ln|x-2|+16ln|x+3|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x^4+x-3)/(2x^3+2x^2-12x)

\int \frac{6 x ^{4} + x - 3}{2 x ^{3} + 2 x ^{2} - 12 x} d x

\frac{3 x ^{2}}{2} - 3 x + \frac{1}{4} ln ∣ x ∣ + \frac{19}{4} ln ∣ x - 2 ∣ + 16 ln ∣ x + 3 ∣ + C

Solution steps

\int \frac{6 x ^{4} + x - 3}{2 x ^{3} + 2 x ^{2} - 12 x} d x

Take the partial fraction of

\frac{6 x ^{4} + x - 3}{2 x ^{3} + 2 x ^{2} - 12 x} : 3 x - 3 + \frac{1}{4 x} + \frac{19}{4 ( x - 2 )} + \frac{16}{x + 3}

= \int 3 x - 3 + \frac{1}{4 x} + \frac{19}{4 ( x - 2 )} + \frac{16}{x + 3} 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 3 x d x - \int 3 d x + \int \frac{1}{4 x} d x + \int \frac{19}{4 ( x - 2 )} d x + \int \frac{16}{x + 3} d x

\int 3 x d x = \frac{3 x ^{2}}{2}

\int 3 d x = 3 x

\int \frac{1}{4 x} d x = \frac{1}{4} ln ∣ x ∣

\int \frac{19}{4 ( x - 2 )} d x = \frac{19}{4} ln ∣ x - 2 ∣

\int \frac{16}{x + 3} d x = 16 ln ∣ x + 3 ∣

= \frac{3 x ^{2}}{2} - 3 x + \frac{1}{4} ln ∣ x ∣ + \frac{19}{4} ln ∣ x - 2 ∣ + 16 ln ∣ x + 3 ∣

Add a constant to the solution

= \frac{3 x ^{2}}{2} - 3 x + \frac{1}{4} ln ∣ x ∣ + \frac{19}{4} ln ∣ x - 2 ∣ + 16 ln ∣ x + 3 ∣ + C

\int \frac{7 x ^{2}}{( x ^{2} + 3 ) ^{2}} d x

\int \frac{10}{x ^{3} - 4 x ^{2} + 8 x} d x

\int (x + 2) 6 - x d x

\int cos (30 x) sin (20 x) d x

\int \frac{e ^{2 x} - 4}{e ^{2 x}} d x

What is the integral of (6x^4+x-3)/(2x^3+2x^2-12x) ?
The integral of (6x^4+x-3)/(2x^3+2x^2-12x) is (3x^2)/2-3x+1/4 ln|x|+19/4 ln|x-2|+16ln|x+3|+C