What is the integral of (4+x)/(sqrt(16-x^2)) ?

The integral of (4+x)/(sqrt(16-x^2)) is -sqrt(16-x^2)+4arcsin(1/4 x)+C

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integral of (4+x)/(sqrt(16-x^2))

\int \frac{4 + x}{16 - x ^{2}} d x

- 16 - x^{2} + 4 arcsin (\frac{1}{4} x) + C

Solution steps

\int \frac{4 + x}{16 - x ^{2}} d x

Apply Trigonometric Substitution

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

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

Take the constant out:

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

= 4 \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

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

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

\int 1 d u = u

= 4 (- cos (u) + u)

Substitute back

u = arcsin (\frac{1}{4} x)

= 4 (- cos (arcsin (\frac{1}{4} x)) + arcsin (\frac{1}{4} x))

Simplify

4 (- cos (arcsin (\frac{1}{4} x)) + arcsin (\frac{1}{4} x)) : - 16 - x^{2} + 4 arcsin (\frac{1}{4} x)

= - 16 - x^{2} + 4 arcsin (\frac{1}{4} x)

Add a constant to the solution

= - 16 - x^{2} + 4 arcsin (\frac{1}{4} x) + C

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

t^{2} y^{^{''}} + t y^{^{'}} + 25 y = 0

\frac{d y}{d x} = 3 x

x \to - 1 lim (- x^{2} + 1)

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 3

What is the integral of (4+x)/(sqrt(16-x^2)) ?
The integral of (4+x)/(sqrt(16-x^2)) is -sqrt(16-x^2)+4arcsin(1/4 x)+C