What is the integral of 1/x-3/(x^2+1) ?

The integral of 1/x-3/(x^2+1) is ln|x|-3arctan(x)+C

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integral of 1/x-3/(x^2+1)

\int \frac{1}{x} - \frac{3}{x ^{2} + 1} d x

ln ∣ x ∣ - 3 arctan (x) + C

Solution steps

\int \frac{1}{x} - \frac{3}{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{3}{x ^{2} + 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{3}{x ^{2} + 1} d x = 3 arctan (x)

= ln ∣ x ∣ - 3 arctan (x)

Add a constant to the solution

= ln ∣ x ∣ - 3 arctan (x) + C

\int (2 sin (x)) d x

\frac{\partial}{\partial x} (e^{x} (x^{2} - y^{2}))

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

\frac{d}{d x} (\frac{3 x ^{3}}{2})

\frac{d}{d x} (sin (x) + x^{2})

What is the integral of 1/x-3/(x^2+1) ?
The integral of 1/x-3/(x^2+1) is ln|x|-3arctan(x)+C