What is the integral of 1/x-4/(x^2+1) ?

The integral of 1/x-4/(x^2+1) is ln|x|-4arctan(x)+C

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integral of 1/x-4/(x^2+1)

\int \frac{1}{x} - \frac{4}{x ^{2} + 1} d x

ln ∣ x ∣ - 4 arctan (x) + C

Solution steps

\int \frac{1}{x} - \frac{4}{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{4}{x ^{2} + 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{4}{x ^{2} + 1} d x = 4 arctan (x)

= ln ∣ x ∣ - 4 arctan (x)

Add a constant to the solution

= ln ∣ x ∣ - 4 arctan (x) + C

\int_{0}^{\infty} cos^{2} (x) d x

\int \frac{1}{64 e ^{- 8 x} + e ^{8 x}} d x

(5, 7), (- 4, - 2)

\int (e^{4 x} + e^{- 4 x})^{2} d x

f (x) = - \frac{10}{x ^{2}}

What is the integral of 1/x-4/(x^2+1) ?
The integral of 1/x-4/(x^2+1) is ln|x|-4arctan(x)+C