What is the integral of (3-18x)/(sqrt(49-9x^2)) ?

The integral of (3-18x)/(sqrt(49-9x^2)) is 2sqrt(-9x^2+49)+arcsin(3/7 x)+C

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integral of (3-18x)/(sqrt(49-9x^2))

\int \frac{3 - 18 x}{49 - 9 x ^{2}} d x

2 - 9 x^{2} + 49 + arcsin (\frac{3}{7} x) + C

Solution steps

\int \frac{3 - 18 x}{49 - 9 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 14 sin (u) + 1 d u

= \int - 14 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 14 sin (u) d u + \int 1 d u

\int 14 sin (u) d u = - 14 cos (u)

\int 1 d u = u

= - (- 14 cos (u)) + u

Substitute back

u = arcsin (\frac{3}{7} x)

= - (- 14 cos (arcsin (\frac{3}{7} x))) + arcsin (\frac{3}{7} x)

Simplify

- (- 14 cos (arcsin (\frac{3}{7} x))) + arcsin (\frac{3}{7} x) : 2 - 9 x^{2} + 49 + arcsin (\frac{3}{7} x)

= 2 - 9 x^{2} + 49 + arcsin (\frac{3}{7} x)

Add a constant to the solution

= 2 - 9 x^{2} + 49 + arcsin (\frac{3}{7} x) + C

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

\int x^{2} (e^{- x}) d x

x \to - 1 f lim (x)

g (x) = x^{2}, g (x) = x^{k} 0, 1

\int \frac{y}{y - 1} d y

What is the integral of (3-18x)/(sqrt(49-9x^2)) ?
The integral of (3-18x)/(sqrt(49-9x^2)) is 2sqrt(-9x^2+49)+arcsin(3/7 x)+C