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

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

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

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

\frac{1}{4} (ln \frac{ln ( x )}{2} + 1 - ln \frac{ln ( x )}{2} - 1) + C

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

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

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= \frac{1}{2} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= \frac{1}{2} \frac{ln \frac{l n ( x )}{2} + 1}{2} - \frac{ln \frac{l n ( x )}{2} - 1}{2}

Simplify

\frac{1}{2} \frac{ln \frac{l n ( x )}{2} + 1}{2} - \frac{ln \frac{l n ( x )}{2} - 1}{2} : \frac{1}{4} (ln \frac{ln ( x )}{2} + 1 - ln \frac{ln ( x )}{2} - 1)

= \frac{1}{4} (ln \frac{ln ( x )}{2} + 1 - ln \frac{ln ( x )}{2} - 1)

Add a constant to the solution

= \frac{1}{4} (ln \frac{ln ( x )}{2} + 1 - ln \frac{ln ( x )}{2} - 1) + C

f (x) = \frac{- 9 x}{2 x - 1}

\int \frac{cos ^{3} ( x )}{1 - sin ( x )} d x

\frac{d}{d x} (se xx)

\int_{- 1}^{- 0.5} 1.5 x^{2} d x

(1 - x^{2}) y^{^{''}} - x y^{^{'}} + y = 0

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