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

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

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integral of (x+1)/(sqrt(16-x^2))

\int \frac{x + 1}{16 - x ^{2}} d x

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

Solution steps

\int \frac{x + 1}{16 - x ^{2}} d x

Expand

\frac{x + 1}{16 - x ^{2}} : \frac{x}{16 - x ^{2}} + \frac{1}{16 - x ^{2}}

= \int \frac{x}{16 - x ^{2}} + \frac{1}{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{x}{16 - x ^{2}} d x + \int \frac{1}{16 - x ^{2}} d x

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

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

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

Add a constant to the solution

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

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