What is the integral of 1/x-3/(16x^2+1) ?

The integral of 1/x-3/(16x^2+1) is ln|x|-3/4 arctan(4x)+C

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integral of 1/x-3/(16x^2+1)

\int \frac{1}{x} - \frac{3}{16 x ^{2} + 1} d x

ln ∣ x ∣ - \frac{3}{4} arctan (4 x) + C

Solution steps

\int \frac{1}{x} - \frac{3}{16 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}{16 x ^{2} + 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{3}{16 x ^{2} + 1} d x = \frac{3}{4} arctan (4 x)

= ln ∣ x ∣ - \frac{3}{4} arctan (4 x)

Add a constant to the solution

= ln ∣ x ∣ - \frac{3}{4} arctan (4 x) + C

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What is the integral of 1/x-3/(16x^2+1) ?
The integral of 1/x-3/(16x^2+1) is ln|x|-3/4 arctan(4x)+C