What is the integral of (3-18x)/(sqrt(16-9x^2)) ?

The integral of (3-18x)/(sqrt(16-9x^2)) is 2sqrt(-9x^2+16)+arcsin(3/4 x)+C

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integral of (3-18x)/(sqrt(16-9x^2))

\int \frac{3 - 18 x}{16 - 9 x ^{2}} d x

2 - 9 x^{2} + 16 + arcsin (\frac{3}{4} x) + C

Solution steps

\int \frac{3 - 18 x}{16 - 9 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 8 sin (u) + 1 d u

= \int - 8 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 8 sin (u) d u + \int 1 d u

\int 8 sin (u) d u = - 8 cos (u)

\int 1 d u = u

= - (- 8 cos (u)) + u

Substitute back

u = arcsin (\frac{3}{4} x)

= - (- 8 cos (arcsin (\frac{3}{4} x))) + arcsin (\frac{3}{4} x)

Simplify

- (- 8 cos (arcsin (\frac{3}{4} x))) + arcsin (\frac{3}{4} x) : 2 - 9 x^{2} + 16 + arcsin (\frac{3}{4} x)

= 2 - 9 x^{2} + 16 + arcsin (\frac{3}{4} x)

Add a constant to the solution

= 2 - 9 x^{2} + 16 + arcsin (\frac{3}{4} x) + C

y^{^{''}} + y^{^{'}} - 2 y = e^{- 2 t}

t \to t lim (3 t^{\frac{1}{3}})

\frac{d}{d x} ((2 x + 1)^{5} (x^{3} - x + 1)^{4})

\int 2 - 2 cos (x) d x

4 x^{\frac{5}{4}} + 2 x^{\frac{3}{2}} + 4 x

What is the integral of (3-18x)/(sqrt(16-9x^2)) ?
The integral of (3-18x)/(sqrt(16-9x^2)) is 2sqrt(-9x^2+16)+arcsin(3/4 x)+C