What is the integral of (1/x-5/(x^2+1)) ?

The integral of (1/x-5/(x^2+1)) is ln|x|-5arctan(x)+C

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integral of (1/x-5/(x^2+1))

\int (\frac{1}{x} - \frac{5}{x ^{2} + 1}) d x

ln ∣ x ∣ - 5 arctan (x) + C

Solution steps

\int \frac{1}{x} - \frac{5}{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{5}{x ^{2} + 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{5}{x ^{2} + 1} d x = 5 arctan (x)

= ln ∣ x ∣ - 5 arctan (x)

Add a constant to the solution

= ln ∣ x ∣ - 5 arctan (x) + C

\frac{s + 2}{s ( s + 2 ) + 1}

\frac{d}{d x} (\frac{3 x + 1}{2 x - 1})

\frac{1}{s ( s + 4 )}

(6 e^{x} cos (x)) (0.6)

sin (\frac{1}{2} x)

What is the integral of (1/x-5/(x^2+1)) ?
The integral of (1/x-5/(x^2+1)) is ln|x|-5arctan(x)+C