What is the integral of (7-98x)/(sqrt(9-49x^2)) ?

The integral of (7-98x)/(sqrt(9-49x^2)) is 2sqrt(-49x^2+9)+arcsin(7/3 x)+C

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integral of (7-98x)/(sqrt(9-49x^2))

\int \frac{7 - 98 x}{9 - 49 x ^{2}} d x

2 - 49 x^{2} + 9 + arcsin (\frac{7}{3} x) + C

Solution steps

\int \frac{7 - 98 x}{9 - 49 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{7}{3} x)

= - (- 6 cos (arcsin (\frac{7}{3} x))) + arcsin (\frac{7}{3} x)

Simplify

- (- 6 cos (arcsin (\frac{7}{3} x))) + arcsin (\frac{7}{3} x) : 2 - 49 x^{2} + 9 + arcsin (\frac{7}{3} x)

= 2 - 49 x^{2} + 9 + arcsin (\frac{7}{3} x)

Add a constant to the solution

= 2 - 49 x^{2} + 9 + arcsin (\frac{7}{3} x) + C

y = 5 + 4 x^{2} - 2 x^{3}

x^{2} + 4 x - 3

3.5 x - 2 x^{2}

\int \frac{1 - 2 x}{x + 1} d x

\frac{d}{d r} (e^{r} + r^{e})

What is the integral of (7-98x)/(sqrt(9-49x^2)) ?
The integral of (7-98x)/(sqrt(9-49x^2)) is 2sqrt(-49x^2+9)+arcsin(7/3 x)+C