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

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

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

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

3 x^{2} - 4 + \frac{x ^{2} - 4}{2 x} + C

Solution steps

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

Expand

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

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

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

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

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

Add a constant to the solution

= 3 x^{2} - 4 + \frac{x ^{2} - 4}{2 x} + C

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

\int sin (2 x + 5) d x

f (x) = arctan (\frac{x - 1}{x + 1})

y = x, at x = 4

f (x) = 9 - x^{2}, at x = - 1

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