What is the integral of (4x-1)/(x^2-25) ?

The integral of (4x-1)/(x^2-25) is 2ln|x^2-25|+1/10 (ln| x/5+1|-ln| x/5-1|)+C

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integral of (4x-1)/(x^2-25)

\int \frac{4 x - 1}{x ^{2} - 25} d x

2 ln x^{2} - 25 + \frac{1}{10} (ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1) + C

Solution steps

\int \frac{4 x - 1}{x ^{2} - 25} d x

Expand

\frac{4 x - 1}{x ^{2} - 25} : \frac{4 x}{x ^{2} - 25} - \frac{1}{x ^{2} - 25}

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

\int \frac{4 x}{x ^{2} - 25} d x = 2 ln x^{2} - 25

\int \frac{1}{x ^{2} - 25} d x = - \frac{1}{10} (ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1)

= 2 ln x^{2} - 25 - (- \frac{1}{10} (ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1))

Simplify

= 2 ln x^{2} - 25 + \frac{1}{10} (ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1)

Add a constant to the solution

= 2 ln x^{2} - 25 + \frac{1}{10} (ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1) + C

5 x^{4} - x^{6}

\int x csc (6 x^{2}) d x

\frac{d y}{d t} = \frac{t e ^{t}}{y 5 + y ^{2}}

\frac{d}{d x} (\frac{sin ( x )}{( 1 + cos ( x ) ) ^{2}})

\frac{d}{d x} (e^{x, x})

What is the integral of (4x-1)/(x^2-25) ?
The integral of (4x-1)/(x^2-25) is 2ln|x^2-25|+1/10 (ln| x/5+1|-ln| x/5-1|)+C