What is the integral of (3x-1)/(xsqrt(x^2-25)) ?

The integral of (3x-1)/(xsqrt(x^2-25)) is 3ln(1/5 |sqrt(x^2-25)+x|)-1/5 arcsec(1/5 x)+C

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integral of (3x-1)/(xsqrt(x^2-25))

\int \frac{3 x - 1}{x x ^{2} - 25} d x

3 ln (\frac{1}{5} x^{2} - 25 + x) - \frac{1}{5} arcsec (\frac{1}{5} x) + C

Solution steps

\int \frac{3 x - 1}{x x ^{2} - 25} d x

Expand

\frac{3 x - 1}{x x ^{2} - 25} : \frac{3}{x ^{2} - 25} - \frac{1}{x x ^{2} - 25}

= \int \frac{3}{x ^{2} - 25} - \frac{1}{x x ^{2} - 25} 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} - 25} d x - \int \frac{1}{x x ^{2} - 25} d x

\int \frac{3}{x ^{2} - 25} d x = 3 ln (\frac{1}{5} x^{2} - 25 + x)

\int \frac{1}{x x ^{2} - 25} d x = \frac{1}{5} arcsec (\frac{1}{5} x)

= 3 ln (\frac{1}{5} x^{2} - 25 + x) - \frac{1}{5} arcsec (\frac{1}{5} x)

Add a constant to the solution

= 3 ln (\frac{1}{5} x^{2} - 25 + x) - \frac{1}{5} arcsec (\frac{1}{5} x) + C

\frac{d}{d x} (ln (3 x - 2))

\frac{d}{d x} (\frac{2}{x ^{3} ln ( 10 )})

\frac{\partial}{\partial y} (y - x)

x = y, y^{2} - 2 y = x

y = 3 x^{2} + 2

What is the integral of (3x-1)/(xsqrt(x^2-25)) ?
The integral of (3x-1)/(xsqrt(x^2-25)) is 3ln(1/5 |sqrt(x^2-25)+x|)-1/5 arcsec(1/5 x)+C