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

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

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

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

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

Solution steps

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

Expand

\frac{4 - x}{16 - x ^{2}} : \frac{4}{16 - x ^{2}} - \frac{x}{16 - x ^{2}}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int \frac{x}{16 - x ^{2}} d x = - 16 - x^{2}

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

Simplify

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

Add a constant to the solution

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

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