What is the integral of 3/(xsqrt(4-9ln^2(x))) ?

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

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integral of 3/(xsqrt(4-9ln^2(x)))

\int \frac{3}{x 4 - 9 ln ^{2} ( x )} d x

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

Solution steps

\int \frac{3}{x 4 - 9 ln ^{2} ( x )} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{x 4 - 9 ln ^{2} ( x )} d x

Apply u-substitution

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

= 3 \cdot \int \frac{1}{4 - 9 u ^{2}} d u

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{4 x ^{\frac{3}{2}}}{x}

y^{^{''}} + 4 y^{^{'}} + 4 y = - (18 t e - 2 t + 8 t + 4)

f (x) = x^{2} - 10 x

\frac{\partial}{\partial x} (5 x^{7} cos (y^{8}))

x \to 0 lim (1 + sin (x) - e^{x})

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