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

The integral of ((2x^2+x+4))/(x^3+4x^2) is 2ln|x+4|-1/x+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+x+4))/(x^3+4x^2)

\int \frac{( 2 x ^{2} + x + 4 )}{x ^{3} + 4 x ^{2}} d x

2 ln ∣ x + 4 ∣ - \frac{1}{x} + C

Solution steps

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

Expand

\frac{2 x ^{2} + x + 4}{x ^{3} + 4 x ^{2}} : \frac{2 x ^{2}}{x ^{3} + 4 x ^{2}} + \frac{x}{x ^{3} + 4 x ^{2}} + \frac{4}{x ^{3} + 4 x ^{2}}

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

\int \frac{2 x ^{2}}{x ^{3} + 4 x ^{2}} d x = 2 ln ∣ x + 4 ∣

\int \frac{x}{x ^{3} + 4 x ^{2}} d x = \frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣

\int \frac{4}{x ^{3} + 4 x ^{2}} d x = - \frac{1}{4} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{4} ln ∣ x + 4 ∣

= 2 ln ∣ x + 4 ∣ + \frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣ - \frac{1}{4} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{4} ln ∣ x + 4 ∣

Simplify

2 ln ∣ x + 4 ∣ + \frac{1}{4} ln ∣ x ∣ - \frac{1}{4} ln ∣ x + 4 ∣ - \frac{1}{4} ln ∣ x ∣ - \frac{1}{x} + \frac{1}{4} ln ∣ x + 4 ∣ : 2 ln ∣ x + 4 ∣ - \frac{1}{x}

= 2 ln ∣ x + 4 ∣ - \frac{1}{x}

Add a constant to the solution

= 2 ln ∣ x + 4 ∣ - \frac{1}{x} + C

\frac{d}{d x} (x^{3} + 6)

y^{'} (x)

13 x y^{^{'}} - 2 y = - \frac{x ^{2}}{y ^{12}}

\frac{d}{d x} (5^{(6 x^{2} - x + 2)})

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

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