What is the integral of (5-50x)/(sqrt(9-25x^2)) ?

The integral of (5-50x)/(sqrt(9-25x^2)) is 2sqrt(-25x^2+9)+arcsin(5/3 x)+C

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integral of (5-50x)/(sqrt(9-25x^2))

\int \frac{5 - 50 x}{9 - 25 x ^{2}} d x

2 - 25 x^{2} + 9 + arcsin (\frac{5}{3} x) + C

Solution steps

\int \frac{5 - 50 x}{9 - 25 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 6 sin (u) + 1 d u

= \int - 6 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 6 sin (u) d u + \int 1 d u

\int 6 sin (u) d u = - 6 cos (u)

\int 1 d u = u

= - (- 6 cos (u)) + u

Substitute back

u = arcsin (\frac{5}{3} x)

= - (- 6 cos (arcsin (\frac{5}{3} x))) + arcsin (\frac{5}{3} x)

Simplify

- (- 6 cos (arcsin (\frac{5}{3} x))) + arcsin (\frac{5}{3} x) : 2 - 25 x^{2} + 9 + arcsin (\frac{5}{3} x)

= 2 - 25 x^{2} + 9 + arcsin (\frac{5}{3} x)

Add a constant to the solution

= 2 - 25 x^{2} + 9 + arcsin (\frac{5}{3} x) + C

\frac{d}{d x} (3 x sin (x^{2}))

f (x) = 2 - 3 x

\int - 7 sec^{2} (x) d x

y^{^{'}} + 5 y = 0, y (0) = 2

\int (5 x^{2} - 5 x + 4 tan (x)) d x

What is the integral of (5-50x)/(sqrt(9-25x^2)) ?
The integral of (5-50x)/(sqrt(9-25x^2)) is 2sqrt(-25x^2+9)+arcsin(5/3 x)+C