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

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

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

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

\frac{1}{2} arcsin (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 Trigonometric Substitution

: \int \frac{1}{2} d v

= \int \frac{1}{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 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{2} \cdot 1 \cdot arcsin (2 ln (x))

Simplify

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

Add a constant to the solution

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

f (x) = \frac{20 x}{x ^{2} - 5}, at x = 5

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

\frac{d ^{3}}{d x ^{3}} (2 x + 8)

\int_{- 5}^{0} (25 - x^{2}) d x

y^{^{'}} = e^{- x}

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