What is the integral of (6x^2+8)/(x^3+4x) ?

The integral of (6x^2+8)/(x^3+4x) is 2ln|x^2+4|+2ln|x|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (6x^2+8)/(x^3+4x)

\int \frac{6 x ^{2} + 8}{x ^{3} + 4 x} d x

2 ln x^{2} + 4 + 2 ln ∣ x ∣ + C

Solution steps

\int \frac{6 x ^{2} + 8}{x ^{3} + 4 x} d x

Expand

\frac{6 x ^{2} + 8}{x ^{3} + 4 x} : \frac{6 x ^{2}}{x ^{3} + 4 x} + \frac{8}{x ^{3} + 4 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{6 x ^{2}}{x ^{3} + 4 x} d x + \int \frac{8}{x ^{3} + 4 x} d x

\int \frac{6 x ^{2}}{x ^{3} + 4 x} d x = 3 ln x^{2} + 4

\int \frac{8}{x ^{3} + 4 x} d x = 2 ln ∣ x ∣ - ln x^{2} + 4

= 3 ln x^{2} + 4 + 2 ln ∣ x ∣ - ln x^{2} + 4

Simplify

= 2 ln x^{2} + 4 + 2 ln ∣ x ∣

Add a constant to the solution

= 2 ln x^{2} + 4 + 2 ln ∣ x ∣ + C

\frac{d y}{d x} + \frac{1}{x} y = x y^{2}

\frac{d y}{d x} + y = e^{7 x}

y^{^{'}} - \frac{5}{x} y = \frac{y ^{5}}{x ^{18}}

\int_{9}^{16} \frac{1}{7 - x} d x

\int \frac{x}{x - 16} d x

What is the integral of (6x^2+8)/(x^3+4x) ?
The integral of (6x^2+8)/(x^3+4x) is 2ln|x^2+4|+2ln|x|+C