What is the integral of (7-98x)/(sqrt(25-49x^2)) ?

The integral of (7-98x)/(sqrt(25-49x^2)) is 2sqrt(-49x^2+25)+arcsin(7/5 x)+C

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integral of (7-98x)/(sqrt(25-49x^2))

\int \frac{7 - 98 x}{25 - 49 x ^{2}} d x

2 - 49 x^{2} + 25 + arcsin (\frac{7}{5} x) + C

Solution steps

\int \frac{7 - 98 x}{25 - 49 x ^{2}} d x

Apply Trigonometric Substitution

: \int - 10 sin (u) + 1 d u

= \int - 10 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 10 sin (u) d u + \int 1 d u

\int 10 sin (u) d u = - 10 cos (u)

\int 1 d u = u

= - (- 10 cos (u)) + u

Substitute back

u = arcsin (\frac{7}{5} x)

= - (- 10 cos (arcsin (\frac{7}{5} x))) + arcsin (\frac{7}{5} x)

Simplify

- (- 10 cos (arcsin (\frac{7}{5} x))) + arcsin (\frac{7}{5} x) : 2 - 49 x^{2} + 25 + arcsin (\frac{7}{5} x)

= 2 - 49 x^{2} + 25 + arcsin (\frac{7}{5} x)

Add a constant to the solution

= 2 - 49 x^{2} + 25 + arcsin (\frac{7}{5} x) + C

\frac{t d u}{d t} = t^{2} + 3 u, H (4) = 48

\frac{9}{x ^{4}}

\int tan (\frac{x}{2}) d x

\frac{\partial}{\partial x} (x^{2} + y cos (x y))

\int 4 sec^{4} (x) tan (x) d x

What is the integral of (7-98x)/(sqrt(25-49x^2)) ?
The integral of (7-98x)/(sqrt(25-49x^2)) is 2sqrt(-49x^2+25)+arcsin(7/5 x)+C