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

\int \frac{cos ( x )}{1 + tan ( x )} d x

\int (5 - 6 x) d x

\int x^{e} + e^{2 x} + x^{π} + π^{x} + 1 d x

\int e^{\frac{- y}{x}} d y

\int 5 x x d 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