What is the integral of 3/(sqrt(5-9x^2)) ?

The integral of 3/(sqrt(5-9x^2)) is arcsin(3/(sqrt(5))x)+C

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integral of 3/(sqrt(5-9x^2))

\int \frac{3}{5 - 9 x ^{2}} d x

arcsin (\frac{3}{5} x) + C

Solution steps

\int \frac{3}{5 - 9 x ^{2}} 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}{5 - 9 x ^{2}} d x

Apply Trigonometric Substitution

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

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

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 u

Integral of a constant:

\int a d x = a x

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

Substitute back

u = arcsin (\frac{3}{5} x)

= 3 \cdot \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{5} x)

Simplify

3 \cdot \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{5} x) : arcsin (\frac{3}{5} x)

= arcsin (\frac{3}{5} x)

Add a constant to the solution

= arcsin (\frac{3}{5} x) + C

\int_{7}^{9} \frac{1}{x ^{2} - 25} d x

f (x) = \frac{1}{10} + x e^{- x}

\frac{0.6 t ^{2} - t + 3}{t ^{2} + 3}

\frac{d}{d x} (8 x - 1)

10 x y y^{^{'}} = 1 - y^{2}

What is the integral of 3/(sqrt(5-9x^2)) ?
The integral of 3/(sqrt(5-9x^2)) is arcsin(3/(sqrt(5))x)+C