What is the integral of (3-18x)/(sqrt(4-9x^2)) ?

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

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integral of (3-18x)/(sqrt(4-9x^2))

\int \frac{3 - 18 x}{4 - 9 x ^{2}} d x

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

Solution steps

\int \frac{3 - 18 x}{4 - 9 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 4 sin (u) + 1 d u

= \int - 4 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 4 sin (u) d u + \int 1 d u

\int 4 sin (u) d u = - 4 cos (u)

\int 1 d u = u

= - (- 4 cos (u)) + u

Substitute back

u = arcsin (\frac{3}{2} x)

= - (- 4 cos (arcsin (\frac{3}{2} x))) + arcsin (\frac{3}{2} x)

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{( 1 - 6 x ) ^{4}} d x

\frac{( s + 1 )}{( s ^{2} + s + 1 )}

x \to + 4 lim (5)

\int_{1}^{2} \frac{2 ln ( x )}{x ^{2}} d x

\frac{d}{d t} (tan (sec (cos (t))))

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