What is the integral of (8+x)/(sqrt(64-x^2)) ?

The integral of (8+x)/(sqrt(64-x^2)) is -sqrt(64-x^2)+8arcsin(1/8 x)+C

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integral of (8+x)/(sqrt(64-x^2))

\int \frac{8 + x}{64 - x ^{2}} d x

- 64 - x^{2} + 8 arcsin (\frac{1}{8} x) + C

Solution steps

\int \frac{8 + x}{64 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 8 (sin (u) + 1) d u

= \int 8 (sin (u) + 1) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int 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

= 8 (\int sin (u) d u + \int 1 d u)

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

\int 1 d u = u

= 8 (- cos (u) + u)

Substitute back

u = arcsin (\frac{1}{8} x)

= 8 (- cos (arcsin (\frac{1}{8} x)) + arcsin (\frac{1}{8} x))

Simplify

8 (- cos (arcsin (\frac{1}{8} x)) + arcsin (\frac{1}{8} x)) : - 64 - x^{2} + 8 arcsin (\frac{1}{8} x)

= - 64 - x^{2} + 8 arcsin (\frac{1}{8} x)

Add a constant to the solution

= - 64 - x^{2} + 8 arcsin (\frac{1}{8} x) + C

\frac{d}{d x} (2 x (cos (x^{2})))

\frac{d}{d x} (x^{3} - 12 x + 2)

\int 4 + 8 cos (θ) + 4 cos^{2} (θ) d θ

x^{2} + y^{^{'}} = x^{2}

\frac{\partial}{\partial x} (6 x e^{y})

What is the integral of (8+x)/(sqrt(64-x^2)) ?
The integral of (8+x)/(sqrt(64-x^2)) is -sqrt(64-x^2)+8arcsin(1/8 x)+C