What is the integral of 1/(3+sqrt(9-x^2)) ?

The integral of 1/(3+sqrt(9-x^2)) is -3/x+arcsin(1/3 x)+(sqrt(9-x^2))/x+C

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integral of 1/(3+sqrt(9-x^2))

\int \frac{1}{3 + 9 - x ^{2}} d x

- \frac{3}{x} + arcsin (\frac{1}{3} x) + \frac{9 - x ^{2}}{x} + C

Solution steps

\int \frac{1}{3 + 9 - x ^{2}} d x

\frac{1}{3 + 9 - x ^{2}} = \frac{3 - 9 - x ^{2}}{( 3 ) ^{2} - ( 9 - x ^{2} ) ^{2}}

= \int \frac{3 - 9 - x ^{2}}{3 ^{2} - ( 9 - x ^{2} ) ^{2}} d x

Expand

\frac{3 - 9 - x ^{2}}{3 ^{2} - ( 9 - x ^{2} ) ^{2}} : \frac{3}{3 ^{2} - ( 9 - x ^{2} ) ^{2}} - \frac{9 - x ^{2}}{3 ^{2} - ( 9 - x ^{2} ) ^{2}}

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{3}{3 ^{2} - ( 9 - x ^{2} ) ^{2}} d x - \int \frac{9 - x ^{2}}{3 ^{2} - ( 9 - x ^{2} ) ^{2}} d x

\int \frac{3}{3 ^{2} - ( 9 - x ^{2} ) ^{2}} d x = - \frac{3}{x}

\int \frac{9 - x ^{2}}{3 ^{2} - ( 9 - x ^{2} ) ^{2}} d x = - arcsin (\frac{1}{3} x) - \frac{9 - x ^{2}}{x}

= - \frac{3}{x} - (- arcsin (\frac{1}{3} x) - \frac{9 - x ^{2}}{x})

Simplify

= - \frac{3}{x} + arcsin (\frac{1}{3} x) + \frac{9 - x ^{2}}{x}

Add a constant to the solution

= - \frac{3}{x} + arcsin (\frac{1}{3} x) + \frac{9 - x ^{2}}{x} + C

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What is the integral of 1/(3+sqrt(9-x^2)) ?
The integral of 1/(3+sqrt(9-x^2)) is -3/x+arcsin(1/3 x)+(sqrt(9-x^2))/x+C