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

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

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

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

- ln 1 + 4 I n^{2} x + ln (4 ∣ x ∣ ∣ I ∣) + 2 ln (n) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( u - 1 )} d u

= \int \frac{1}{u ( u - 1 )} d u

Take the partial fraction of

\frac{1}{u ( u - 1 )} : - \frac{1}{u} + \frac{1}{u - 1}

= \int - \frac{1}{u} + \frac{1}{u - 1} d u

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}{u} d u + \int \frac{1}{u - 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

= - ln ∣ u ∣ + ln ∣ u - 1 ∣

Substitute back

u = 1 + 4 I n^{2} x

= - ln 1 + 4 I n^{2} x + ln 1 + 4 I n^{2} x - 1

Simplify

- ln 1 + 4 I n^{2} x + ln 1 + 4 I n^{2} x - 1 : - ln 1 + 4 I n^{2} x + ln (4 ∣ x ∣ ∣ I ∣) + 2 ln (n)

= - ln 1 + 4 I n^{2} x + ln (4 ∣ x ∣ ∣ I ∣) + 2 ln (n)

Add a constant to the solution

= - ln 1 + 4 I n^{2} x + ln (4 ∣ x ∣ ∣ I ∣) + 2 ln (n) + C

y = 16 x

f (v) = \frac{v}{v + \frac{d}{v}}

y = 3 x, y = \frac{3}{2} x

3 e^{- s}

\int_{1}^{\infty} 8 e^{- 4 x} d x

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