What is the integral of (6-72x)/(sqrt(49-36x^2)) ?

The integral of (6-72x)/(sqrt(49-36x^2)) is 2sqrt(-36x^2+49)+arcsin(6/7 x)+C

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integral of (6-72x)/(sqrt(49-36x^2))

\int \frac{6 - 72 x}{49 - 36 x ^{2}} d x

2 - 36 x^{2} + 49 + arcsin (\frac{6}{7} x) + C

Solution steps

\int \frac{6 - 72 x}{49 - 36 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{6}{7} x)

= - (- 14 cos (arcsin (\frac{6}{7} x))) + arcsin (\frac{6}{7} x)

Simplify

- (- 14 cos (arcsin (\frac{6}{7} x))) + arcsin (\frac{6}{7} x) : 2 - 36 x^{2} + 49 + arcsin (\frac{6}{7} x)

= 2 - 36 x^{2} + 49 + arcsin (\frac{6}{7} x)

Add a constant to the solution

= 2 - 36 x^{2} + 49 + arcsin (\frac{6}{7} x) + C

f (x) = (x - 4) (x^{2} + 8), (1, - 27)

x \to 1 - lim (- x^{2} + 5 x + (- 4))

n = 3 \sum \infty \frac{2}{n ln ^{2} ( n )}

y = x, y = 3 x

\int \frac{x}{( 5 x ^{2} + 2 ) ^{6}} d x

What is the integral of (6-72x)/(sqrt(49-36x^2)) ?
The integral of (6-72x)/(sqrt(49-36x^2)) is 2sqrt(-36x^2+49)+arcsin(6/7 x)+C