What is the integral of 1/x-6/(x^2+1) ?

The integral of 1/x-6/(x^2+1) is ln|x|-6arctan(x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/x-6/(x^2+1)

\int \frac{1}{x} - \frac{6}{x ^{2} + 1} d x

ln ∣ x ∣ - 6 arctan (x) + C

Solution steps

\int \frac{1}{x} - \frac{6}{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{6}{x ^{2} + 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{6}{x ^{2} + 1} d x = 6 arctan (x)

= ln ∣ x ∣ - 6 arctan (x)

Add a constant to the solution

= ln ∣ x ∣ - 6 arctan (x) + C

f (x) = sin (cos (x^{2}))

n = 1 \sum \infty 3 n^{5}

\frac{d}{d x} (ln (x^{2} + 1) - ln (x + 1))

\frac{\partial}{\partial z} (3 x sin (y - z))

y^{^{''}} + 5 y^{^{'}} - 14 y = e^{(4 x)}

What is the integral of 1/x-6/(x^2+1) ?
The integral of 1/x-6/(x^2+1) is ln|x|-6arctan(x)+C