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

x \to 1 lim (- 2)

4 x - 7

\int \frac{5 x ^{2}}{x ^{2} + 8} d x

y = 3 + 7 - x, (3, 5)

y = 2 x^{2} - 3 x + 1, (3, 10)

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