What is the integral of ((x+3))/(sqrt(16-x^2)) ?

The integral of ((x+3))/(sqrt(16-x^2)) is -sqrt(16-x^2)+3arcsin(1/4 x)+C

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integral of ((x+3))/(sqrt(16-x^2))

\int \frac{( x + 3 )}{16 - x ^{2}} d x

- 16 - x^{2} + 3 arcsin (\frac{1}{4} x) + C

Solution steps

\int \frac{x + 3}{16 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 4 sin (u) + 3 d u

= \int 4 sin (u) + 3 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 3 d u

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

\int 3 d u = 3 u

= - 4 cos (u) + 3 u

Substitute back

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

= - 4 cos (arcsin (\frac{1}{4} x)) + 3 arcsin (\frac{1}{4} x)

Simplify

- 4 cos (arcsin (\frac{1}{4} x)) + 3 arcsin (\frac{1}{4} x) : - 16 - x^{2} + 3 arcsin (\frac{1}{4} x)

= - 16 - x^{2} + 3 arcsin (\frac{1}{4} x)

Add a constant to the solution

= - 16 - x^{2} + 3 arcsin (\frac{1}{4} x) + C

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What is the integral of ((x+3))/(sqrt(16-x^2)) ?
The integral of ((x+3))/(sqrt(16-x^2)) is -sqrt(16-x^2)+3arcsin(1/4 x)+C