What is the integral of 9/(xsqrt(4-81ln^2(x))) ?

The integral of 9/(xsqrt(4-81ln^2(x))) is arcsin(9/2 ln(x))+C

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integral of 9/(xsqrt(4-81ln^2(x)))

\int \frac{9}{x 4 - 81 ln ^{2} ( x )} d x

arcsin (\frac{9}{2} ln (x)) + C

Solution steps

\int \frac{9}{x 4 - 81 ln ^{2} ( x )} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{x 4 - 81 ln ^{2} ( x )} d x

Apply u-substitution

: \int \frac{1}{4 - 81 u ^{2}} d u

= 9 \cdot \int \frac{1}{4 - 81 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{9} d v

= 9 \cdot \int \frac{1}{9} d v

Take the constant out:

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

= 9 \cdot \frac{1}{9} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 9 \cdot \frac{1}{9} \cdot 1 \cdot v

Substitute back

= 9 \cdot \frac{1}{9} \cdot 1 \cdot arcsin (\frac{9}{2} ln (x))

Simplify

9 \cdot \frac{1}{9} \cdot 1 \cdot arcsin (\frac{9}{2} ln (x)) : arcsin (\frac{9}{2} ln (x))

= arcsin (\frac{9}{2} ln (x))

Add a constant to the solution

= arcsin (\frac{9}{2} ln (x)) + C

\int 50 - \frac{x}{2} d x

\frac{d x}{d y} = - \frac{x}{y}

\frac{d y}{d x} - e^{6 x + y} = 0

x \to 1 lim (\frac{1}{x ^{2}})

y^{^{''}} - 4 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 3

What is the integral of 9/(xsqrt(4-81ln^2(x))) ?
The integral of 9/(xsqrt(4-81ln^2(x))) is arcsin(9/2 ln(x))+C