What is the integral of (9-162x)/(sqrt(64-81x^2)) ?

The integral of (9-162x)/(sqrt(64-81x^2)) is 2sqrt(-81x^2+64)+arcsin(9/8 x)+C

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integral of (9-162x)/(sqrt(64-81x^2))

\int \frac{9 - 162 x}{64 - 81 x ^{2}} d x

2 - 81 x^{2} + 64 + arcsin (\frac{9}{8} x) + C

Solution steps

\int \frac{9 - 162 x}{64 - 81 x ^{2}} d x

Apply Trigonometric Substitution

= \int - 16 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 16 sin (u) d u + \int 1 d u

\int 16 sin (u) d u = - 16 cos (u)

\int 1 d u = u

= - (- 16 cos (u)) + u

Substitute back

u = arcsin (\frac{9}{8} x)

= - (- 16 cos (arcsin (\frac{9}{8} x))) + arcsin (\frac{9}{8} x)

Simplify

- (- 16 cos (arcsin (\frac{9}{8} x))) + arcsin (\frac{9}{8} x) : 2 - 81 x^{2} + 64 + arcsin (\frac{9}{8} x)

= 2 - 81 x^{2} + 64 + arcsin (\frac{9}{8} x)

Add a constant to the solution

= 2 - 81 x^{2} + 64 + arcsin (\frac{9}{8} x) + C

\int - 2 x \cdot ln (x) d x

t an g e n t x^{2} - \frac{1}{x}

\int_{3}^{6} \frac{2 x - 3}{6} d x

\int_{0}^{π} x \cdot sin^{2} (x) d x

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

What is the integral of (9-162x)/(sqrt(64-81x^2)) ?
The integral of (9-162x)/(sqrt(64-81x^2)) is 2sqrt(-81x^2+64)+arcsin(9/8 x)+C