What is the integral of (7-98x)/(sqrt(4-49x^2)) ?

The integral of (7-98x)/(sqrt(4-49x^2)) is 2sqrt(-49x^2+4)+arcsin(7/2 x)+C

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integral of (7-98x)/(sqrt(4-49x^2))

\int \frac{7 - 98 x}{4 - 49 x ^{2}} d x

2 - 49 x^{2} + 4 + arcsin (\frac{7}{2} x) + C

Solution steps

\int \frac{7 - 98 x}{4 - 49 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 4 sin (u) + 1 d u

= \int - 4 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 4 sin (u) d u + \int 1 d u

\int 4 sin (u) d u = - 4 cos (u)

\int 1 d u = u

= - (- 4 cos (u)) + u

Substitute back

u = arcsin (\frac{7}{2} x)

= - (- 4 cos (arcsin (\frac{7}{2} x))) + arcsin (\frac{7}{2} x)

Simplify

- (- 4 cos (arcsin (\frac{7}{2} x))) + arcsin (\frac{7}{2} x) : 2 - 49 x^{2} + 4 + arcsin (\frac{7}{2} x)

= 2 - 49 x^{2} + 4 + arcsin (\frac{7}{2} x)

Add a constant to the solution

= 2 - 49 x^{2} + 4 + arcsin (\frac{7}{2} x) + C

e^{- 9 x}

x \to 1 + lim (\frac{7}{x - 1})

x ln (x) \frac{d y}{d x} + y = x e^{x}

\int ∣ x - 1 ∣ d x

\int \frac{- 2}{x ^{2} + 16} d x

What is the integral of (7-98x)/(sqrt(4-49x^2)) ?
The integral of (7-98x)/(sqrt(4-49x^2)) is 2sqrt(-49x^2+4)+arcsin(7/2 x)+C