What is the integral of (3-18x)/(sqrt(25-9x^2)) ?

The integral of (3-18x)/(sqrt(25-9x^2)) is 2sqrt(-9x^2+25)+arcsin(3/5 x)+C

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integral of (3-18x)/(sqrt(25-9x^2))

\int \frac{3 - 18 x}{25 - 9 x ^{2}} d x

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

Solution steps

\int \frac{3 - 18 x}{25 - 9 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 10 sin (u) + 1 d u

= \int - 10 sin (u) + 1 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

= - \int 10 sin (u) d u + \int 1 d u

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

\int 1 d u = u

= - (- 10 cos (u)) + u

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

ln (\frac{4 x}{x + 7})

\int \frac{( e ^{x} )}{e ^{2 x} + 3} d x

\int_{0}^{1} x x^{2} + 2 d x

2 \frac{d y}{d x} + y = 0

y = x^{m x}

What is the integral of (3-18x)/(sqrt(25-9x^2)) ?
The integral of (3-18x)/(sqrt(25-9x^2)) is 2sqrt(-9x^2+25)+arcsin(3/5 x)+C