What is the integral of (3-6x)/(sqrt(25-9x^2)) ?

The integral of (3-6x)/(sqrt(25-9x^2)) is arcsin(3/5 x)+2/3 sqrt(-9x^2+25)+C

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integral of (3-6x)/(sqrt(25-9x^2))

\int \frac{3 - 6 x}{25 - 9 x ^{2}} d x

arcsin (\frac{3}{5} x) + \frac{2}{3} - 9 x^{2} + 25 + C

Solution steps

\int \frac{3 - 6 x}{25 - 9 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{3 - 10 sin ( u )}{3} d u

= \int \frac{3 - 10 sin ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int 3 - 10 sin (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int 3 d u - \int 10 sin (u) d u)

\int 3 d u = 3 u

\int 10 sin (u) d u = - 10 cos (u)

= \frac{1}{3} (3 u - (- 10 cos (u)))

Substitute back

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

= \frac{1}{3} (3 arcsin (\frac{3}{5} x) - (- 10 cos (arcsin (\frac{3}{5} x))))

Simplify

\frac{1}{3} (3 arcsin (\frac{3}{5} x) - (- 10 cos (arcsin (\frac{3}{5} x)))) : arcsin (\frac{3}{5} x) + \frac{2}{3} - 9 x^{2} + 25

= arcsin (\frac{3}{5} x) + \frac{2}{3} - 9 x^{2} + 25

Add a constant to the solution

= arcsin (\frac{3}{5} x) + \frac{2}{3} - 9 x^{2} + 25 + C

\frac{\partial}{\partial x} ((x + y) e^{3 x - 4 y})

\frac{d ^{2}}{d x ^{2}} (\frac{e ^{x}}{8 x + 1})

ln (ln (x^{2}))

y^{^{'}} - \frac{5}{x} y = \frac{y ^{5}}{x ^{11}}

n = 1 \sum \infty \frac{4}{1 0 ^{n}}

What is the integral of (3-6x)/(sqrt(25-9x^2)) ?
The integral of (3-6x)/(sqrt(25-9x^2)) is arcsin(3/5 x)+2/3 sqrt(-9x^2+25)+C