What is the integral of (7-98x)/(sqrt(64-49x^2)) ?

The integral of (7-98x)/(sqrt(64-49x^2)) is 2sqrt(-49x^2+64)+arcsin(7/8 x)+C

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integral of (7-98x)/(sqrt(64-49x^2))

\int \frac{7 - 98 x}{64 - 49 x ^{2}} d x

2 - 49 x^{2} + 64 + arcsin (\frac{7}{8} x) + C

Solution steps

\int \frac{7 - 98 x}{64 - 49 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 16 sin (u) + 1 d u

= \int - 16 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 16 sin (u) d u + \int 1 d u

\int 16 sin (u) d u = - 16 cos (u)

\int 1 d u = u

= - (- 16 cos (u)) + u

Substitute back

u = arcsin (\frac{7}{8} x)

= - (- 16 cos (arcsin (\frac{7}{8} x))) + arcsin (\frac{7}{8} x)

Simplify

- (- 16 cos (arcsin (\frac{7}{8} x))) + arcsin (\frac{7}{8} x) : 2 - 49 x^{2} + 64 + arcsin (\frac{7}{8} x)

= 2 - 49 x^{2} + 64 + arcsin (\frac{7}{8} x)

Add a constant to the solution

= 2 - 49 x^{2} + 64 + arcsin (\frac{7}{8} x) + C

\int \frac{x ^{2}}{a ^{2}} d x

\int 2 x \cdot e^{4 x + 2} d x

(t^{2} + t) cos (5 t)

y^{^{''}} + 4 y^{^{'}} + 4 y = - 3 e^{2 t}

n = 1 \sum \infty \frac{1}{2 ^{n + 2}}

What is the integral of (7-98x)/(sqrt(64-49x^2)) ?
The integral of (7-98x)/(sqrt(64-49x^2)) is 2sqrt(-49x^2+64)+arcsin(7/8 x)+C