What is the integral of (1/x-5/(4x^2+1)) ?

The integral of (1/x-5/(4x^2+1)) is ln|x|-5/2 arctan(2x)+C

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integral of (1/x-5/(4x^2+1))

\int (\frac{1}{x} - \frac{5}{4 x ^{2} + 1}) d x

ln ∣ x ∣ - \frac{5}{2} arctan (2 x) + C

Solution steps

\int \frac{1}{x} - \frac{5}{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{5}{4 x ^{2} + 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{5}{4 x ^{2} + 1} d x = \frac{5}{2} arctan (2 x)

= ln ∣ x ∣ - \frac{5}{2} arctan (2 x)

Add a constant to the solution

= ln ∣ x ∣ - \frac{5}{2} arctan (2 x) + C

\int \frac{( 17 )}{( x - 1 ) ( x ^{2} + 16 )} d x

\int x (3 x^{2} - 2) d x

\frac{d}{d x} (\frac{1}{5} x^{5} - 3 x^{4} + 9 x^{2} + 1)

\frac{\partial}{\partial x} (cos (2 x + y) - z)

\frac{d y}{d x} = \frac{6 y}{x}

What is the integral of (1/x-5/(4x^2+1)) ?
The integral of (1/x-5/(4x^2+1)) is ln|x|-5/2 arctan(2x)+C