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

Popular >

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 (x^{2} - 9 x + 5) d x

x \to + 1 lim (x^{2})

\frac{- 6 x ^{2} - 4 x + 4}{( 3 x ^{2} + 2 ) ^{2}}

\int \frac{sin ( y )}{cos ^{2} ( y )} d y

x \to 2 lim (\frac{1}{3})

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