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

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

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integral of (3x^2+1)/(x^2sqrt(4-x^2))

\int \frac{3 x ^{2} + 1}{x ^{2} 4 - x ^{2}} d x

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

Solution steps

\int \frac{3 x ^{2} + 1}{x ^{2} 4 - x ^{2}} d x

Expand

\frac{3 x ^{2} + 1}{x ^{2} 4 - x ^{2}} : \frac{3}{4 - x ^{2}} + \frac{1}{x ^{2} 4 - x ^{2}}

= \int \frac{3}{4 - x ^{2}} + \frac{1}{x ^{2} 4 - 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{3}{4 - x ^{2}} d x + \int \frac{1}{x ^{2} 4 - x ^{2}} d x

\int \frac{3}{4 - x ^{2}} d x = 3 arcsin (\frac{1}{2} x)

\int \frac{1}{x ^{2} 4 - x ^{2}} d x = - \frac{4 - x ^{2}}{4 x}

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{cos ( x )}{x ^{4}})

\frac{d}{d x} (\frac{y ^{2}}{x})

\frac{d}{d x} (ln (\frac{1}{3 x}))

\int sin (3 π) t d t

\frac{d}{d x} (3 e^{- x^{3}})

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