What is the integral of (2x^2+13x+18)/(x^3+6x^2+9x) ?

The integral of (2x^2+13x+18)/(x^3+6x^2+9x) is 2ln|x|-1/(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 (2x^2+13x+18)/(x^3+6x^2+9x)

\int \frac{2 x ^{2} + 13 x + 18}{x ^{3} + 6 x ^{2} + 9 x} d x

2 ln ∣ x ∣ - \frac{1}{x + 3} + C

Solution steps

\int \frac{2 x ^{2} + 13 x + 18}{x ^{3} + 6 x ^{2} + 9 x} d x

Take the partial fraction of

\frac{2 x ^{2} + 13 x + 18}{x ^{3} + 6 x ^{2} + 9 x} : \frac{2}{x} + \frac{1}{( x + 3 ) ^{2}}

= \int \frac{2}{x} + \frac{1}{( x + 3 ) ^{2}} 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{2}{x} d x + \int \frac{1}{( x + 3 ) ^{2}} d x

\int \frac{2}{x} d x = 2 ln ∣ x ∣

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

= 2 ln ∣ x ∣ - \frac{1}{x + 3}

Add a constant to the solution

= 2 ln ∣ x ∣ - \frac{1}{x + 3} + C

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{2 x}

\int 9 x^{2} - 4 x d x

\frac{\partial}{\partial x} (1 + ln (x))

s (t) = \frac{t ^{2} + 40 t}{5}

\frac{\partial}{\partial x} (e^{(x^{2} - y^{2})})

What is the integral of (2x^2+13x+18)/(x^3+6x^2+9x) ?
The integral of (2x^2+13x+18)/(x^3+6x^2+9x) is 2ln|x|-1/(x+3)+C