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

The integral of ((3x-1))/(xsqrt(x^2-4)) is 3ln(1/2 |sqrt(x^2-4)+x|)-1/2 arcsec(1/2 x)+C

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

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

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

Solution steps

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

Expand

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

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

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

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

= 3 ln (\frac{1}{2} x^{2} - 4 + x) - \frac{1}{2} arcsec (\frac{1}{2} x)

Add a constant to the solution

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

y^{^{''}} - y = 8 x e^{x}

x \to 1 lim (7^{x^{4} - x^{2}})

\int_{0}^{1} ln (x + 1) d x

y = \frac{( 4 x ^{2} + 6 x + 8 )}{x}

- 4 e^{- \frac{x}{3}} - 2

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