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

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

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

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

ln ∣ x ∣ - \frac{1}{2} ln 4 x^{2} + 1 + C

Solution steps

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

Take the partial fraction of

\frac{1}{x ( 1 + 4 x ^{2} )} : \frac{1}{x} - \frac{4 x}{4 x ^{2} + 1}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - \frac{1}{2} ln 4 x^{2} + 1

Add a constant to the solution

= ln ∣ x ∣ - \frac{1}{2} ln 4 x^{2} + 1 + C

x \to + 0 - lim (arctan (\frac{1}{x}))

\frac{- a}{x ^{2}} + \frac{a}{( x - 2 ) ^{2}}

f (x) = \frac{2 x + 3}{x ^{2} + 3 x + 15}

\int_{0}^{2} \frac{1}{4 - 2 x} d x

\int_{1}^{5} (2 + 3 x - x^{2}) d x

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