What is the integral of 1/x-3/(x^2+1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/x-3/(x^2+1)

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

ln ∣ x ∣ - 3 arctan (x) + C

Solution steps

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - 3 arctan (x)

Add a constant to the solution

= ln ∣ x ∣ - 3 arctan (x) + C

\int (2 sin (x)) d x

\frac{d}{d x} (\frac{3 x ^{3}}{2})

(y e^{x y} cos (2 x) - 2 e^{x y} sin (2 x) + 2 x) + (x e^{x y} cos (2 x) - 3) y^{^{'}} = 0

x \to 0 lim ((1 - 6 x)^{\frac{5}{x}})

\frac{d y}{d x} - sin (x) y = 2 sin (x)

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