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

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

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

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

\frac{1}{2} arctan (2 ln (x)) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{1 + 4 u ^{2}} d u

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

Apply Integral Substitution

= \int \frac{1}{2 ( 1 + v ^{2} )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

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

= \frac{1}{2} arctan (v)

Substitute back

= \frac{1}{2} arctan (2 ln (x))

Add a constant to the solution

= \frac{1}{2} arctan (2 ln (x)) + C

\frac{d y}{d x} = 2 x + 2 x y

x \to \infty lim (3 x^{5})

y^{2} \frac{d y}{d x} + y^{3} = 1

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

\int_{0}^{7} 245000 π (7 - y) d y

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