What is the integral of (7-98x)/(sqrt(81-49x^2)) ?

The integral of (7-98x)/(sqrt(81-49x^2)) is 2sqrt(-49x^2+81)+arcsin(7/9 x)+C

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integral of (7-98x)/(sqrt(81-49x^2))

\int \frac{7 - 98 x}{81 - 49 x ^{2}} d x

2 - 49 x^{2} + 81 + arcsin (\frac{7}{9} x) + C

Solution steps

\int \frac{7 - 98 x}{81 - 49 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 18 sin (u) + 1 d u

= \int - 18 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 18 sin (u) d u + \int 1 d u

\int 18 sin (u) d u = - 18 cos (u)

\int 1 d u = u

= - (- 18 cos (u)) + u

Substitute back

u = arcsin (\frac{7}{9} x)

= - (- 18 cos (arcsin (\frac{7}{9} x))) + arcsin (\frac{7}{9} x)

Simplify

- (- 18 cos (arcsin (\frac{7}{9} x))) + arcsin (\frac{7}{9} x) : 2 - 49 x^{2} + 81 + arcsin (\frac{7}{9} x)

= 2 - 49 x^{2} + 81 + arcsin (\frac{7}{9} x)

Add a constant to the solution

= 2 - 49 x^{2} + 81 + arcsin (\frac{7}{9} x) + C

y = x^{0.4}

\frac{d}{d x} ((x + 1)^{3} (x - 2))

\frac{d}{d x} (9 x + 5)

\int sec (32 x) tan (32 x) d x

\frac{1}{x + 2}

What is the integral of (7-98x)/(sqrt(81-49x^2)) ?
The integral of (7-98x)/(sqrt(81-49x^2)) is 2sqrt(-49x^2+81)+arcsin(7/9 x)+C