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

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

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

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

ln ∣ x ∣ - 2 arctan (x) + C

Solution steps

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - 2 arctan (x)

Add a constant to the solution

= ln ∣ x ∣ - 2 arctan (x) + C

\frac{( - 2 )}{( 6 s + s ^{2} )}

y d x = 2 (x + y) d y

(x + y) d x + (x + 2 y) d y = 0, y (2) = 3

x \to 0 - lim (\frac{1}{1 + e ^{\frac{1}{x}}})

\frac{d}{d x} (2 arctan (4 x))

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