What is the integral of (9-162x)/(sqrt(4-81x^2)) ?

The integral of (9-162x)/(sqrt(4-81x^2)) is 2sqrt(-81x^2+4)+arcsin(9/2 x)+C

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integral of (9-162x)/(sqrt(4-81x^2))

\int \frac{9 - 162 x}{4 - 81 x ^{2}} d x

2 - 81 x^{2} + 4 + arcsin (\frac{9}{2} x) + C

Solution steps

\int \frac{9 - 162 x}{4 - 81 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{9}{2} x)

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

Simplify

- (- 4 cos (arcsin (\frac{9}{2} x))) + arcsin (\frac{9}{2} x) : 2 - 81 x^{2} + 4 + arcsin (\frac{9}{2} x)

= 2 - 81 x^{2} + 4 + arcsin (\frac{9}{2} x)

Add a constant to the solution

= 2 - 81 x^{2} + 4 + arcsin (\frac{9}{2} x) + C

\int_{1}^{8} \frac{x}{x ^{2} + 4 x + 20} d x

\frac{1}{x ^{4}}

\int_{0}^{4} (4 e^{- 0.25 x} - 6 x) d x

\frac{x}{lo g _{e} ( x )}

5 - 2 x

What is the integral of (9-162x)/(sqrt(4-81x^2)) ?
The integral of (9-162x)/(sqrt(4-81x^2)) is 2sqrt(-81x^2+4)+arcsin(9/2 x)+C