What is the integral of 1/(xsqrt(4-9ln^2(x))) ?

The integral of 1/(xsqrt(4-9ln^2(x))) is 1/3 arcsin(3/2 ln(x))+C

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integral of 1/(xsqrt(4-9ln^2(x)))

\int \frac{1}{x 4 - 9 ln ^{2} ( x )} d x

\frac{1}{3} arcsin (\frac{3}{2} ln (x)) + C

Solution steps

\int \frac{1}{x 4 - 9 ln ^{2} ( x )} d x

Apply u-substitution

= \int \frac{1}{4 - 9 u ^{2}} d u

Apply Trigonometric Substitution

= \int \frac{1}{3} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{2} ln (x))

Simplify

= \frac{1}{3} arcsin (\frac{3}{2} ln (x))

Add a constant to the solution

= \frac{1}{3} arcsin (\frac{3}{2} ln (x)) + C

What is the integral of 1/(xsqrt(4-9ln^2(x))) ?
The integral of 1/(xsqrt(4-9ln^2(x))) is 1/3 arcsin(3/2 ln(x))+C