What is the integral of 2/(1+x^2)-4/x ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2/(1+x^2)-4/x

\int \frac{2}{1 + x ^{2}} - \frac{4}{x} d x

2 arctan (x) - 4 ln ∣ x ∣ + C

Solution steps

\int \frac{2}{1 + x ^{2}} - \frac{4}{x} 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}{1 + x ^{2}} d x - \int \frac{4}{x} d x

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

\int \frac{4}{x} d x = 4 ln ∣ x ∣

= 2 arctan (x) - 4 ln ∣ x ∣

Add a constant to the solution

= 2 arctan (x) - 4 ln ∣ x ∣ + C

y = 25 - x^{2}

x + 1, 0

\int k e^{x} d x

n = 1 \sum \infty n \cdot e^{- n^{2}}

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

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