What is the integral of 6/(xsqrt(36-(ln(x))^2)) ?

The integral of 6/(xsqrt(36-(ln(x))^2)) is 6arcsin(1/6 ln(x))+C

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integral of 6/(xsqrt(36-(ln(x))^2))

\int \frac{6}{x 36 - ( ln ( x ) ) ^{2}} d x

6 arcsin (\frac{1}{6} ln (x)) + C

Solution steps

\int \frac{6}{x 36 - ( ln ( x ) ) ^{2}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{x 36 - ( ln ( x ) ) ^{2}} d x

Simplify

= 6 \cdot \int \frac{1}{x 36 - ln ^{2} ( x )} d x

Apply u-substitution

: \int \frac{1}{36 - u ^{2}} d u

= 6 \cdot \int \frac{1}{36 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

= 6 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 6 \cdot 1 \cdot v

Substitute back

= 6 \cdot 1 \cdot arcsin (\frac{1}{6} ln (x))

Simplify

= 6 arcsin (\frac{1}{6} ln (x))

Add a constant to the solution

= 6 arcsin (\frac{1}{6} ln (x)) + C

f (x) = 6 x cos (x), (π, - 6 π)

\frac{d}{d x} (e^{2 x} - e^{- 2 x} - 4 x)

\frac{3}{x + 9}

y \cdot y^{^{'}} = \frac{5 x}{4 x ^{2} + 1}

What is the integral of 6/(xsqrt(36-(ln(x))^2)) ?
The integral of 6/(xsqrt(36-(ln(x))^2)) is 6arcsin(1/6 ln(x))+C