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

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

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

\int \frac{3 x ^{2} + 1}{x ^{2} \cdot 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{\partial}{\partial x} (sin (x \cdot y \cdot e^{x^{2}}))

f^{^{'}} (x) = 2^{4 x} + 4 x^{2}

\frac{\partial}{\partial x} (x^{4} + y^{4})

x \to + 0 lim (- x^{2} + 5 x + a)

h \to 0 lim (\frac{( x + h ) - x}{h})

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