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

The integral of (x^2-2x+3)/(x^3+3x) is ln|x|-2/(sqrt(3))arctan(x/(sqrt(3)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2-2x+3)/(x^3+3x)

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

ln ∣ x ∣ - \frac{2}{3} arctan (\frac{x}{3}) + C

Solution steps

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

Take the partial fraction of

\frac{x ^{2} - 2 x + 3}{x ^{3} + 3 x} : \frac{1}{x} - \frac{2}{x ^{2} + 3}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{2}{3} arctan (\frac{x}{3})

Add a constant to the solution

= ln ∣ x ∣ - \frac{2}{3} arctan (\frac{x}{3}) + C

\frac{\partial}{\partial x} (\frac{1}{y ^{2}})

x d x + y e^{- x} d y = 0, y (0) = 1

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 y = 0

y = cos (sin (2 x + 5))

\int sec^{2} (7 x) d x

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