What is the integral of 5/(xsqrt(36-25ln^2(x))) ?

The integral of 5/(xsqrt(36-25ln^2(x))) is arcsin(5/6 ln(x))+C

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integral of 5/(xsqrt(36-25ln^2(x)))

\int \frac{5}{x 36 - 25 ln ^{2} ( x )} d x

arcsin (\frac{5}{6} ln (x)) + C

Solution steps

\int \frac{5}{x 36 - 25 ln ^{2} ( x )} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{x 36 - 25 ln ^{2} ( x )} d x

Apply u-substitution

: \int \frac{1}{36 - 25 u ^{2}} d u

= 5 \cdot \int \frac{1}{36 - 25 u ^{2}} d u

Apply Trigonometric Substitution

: \int \frac{1}{5} d v

= 5 \cdot \int \frac{1}{5} d v

Take the constant out:

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

= 5 \cdot \frac{1}{5} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 5 \cdot \frac{1}{5} \cdot 1 \cdot v

Substitute back

= 5 \cdot \frac{1}{5} \cdot 1 \cdot arcsin (\frac{5}{6} ln (x))

Simplify

5 \cdot \frac{1}{5} \cdot 1 \cdot arcsin (\frac{5}{6} ln (x)) : arcsin (\frac{5}{6} ln (x))

= arcsin (\frac{5}{6} ln (x))

Add a constant to the solution

= arcsin (\frac{5}{6} ln (x)) + C

\frac{\partial}{\partial x} (ln (y^{x^{2}}))

\frac{d}{d x} (\frac{e ^{x}}{x})

\frac{7}{3 x + 4}

t y^{^{'}} + 4 y = 0

\int \frac{1}{y ^{2} + 3} d y

What is the integral of 5/(xsqrt(36-25ln^2(x))) ?
The integral of 5/(xsqrt(36-25ln^2(x))) is arcsin(5/6 ln(x))+C