What is the integral of (7x-1)/(x^2-1) ?

The integral of (7x-1)/(x^2-1) is 7/2 ln|x^2-1|+1/2 ln|x+1|-1/2 ln|x-1|+C

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integral of (7x-1)/(x^2-1)

\int \frac{7 x - 1}{x ^{2} - 1} d x

\frac{7}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + C

Solution steps

\int \frac{7 x - 1}{x ^{2} - 1} d x

Expand

\frac{7 x - 1}{x ^{2} - 1} : \frac{7 x}{x ^{2} - 1} - \frac{1}{x ^{2} - 1}

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{7 x}{x ^{2} - 1} d x - \int \frac{1}{x ^{2} - 1} d x

\int \frac{7 x}{x ^{2} - 1} d x = \frac{7}{2} ln x^{2} - 1

\int \frac{1}{x ^{2} - 1} d x = - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣

= \frac{7}{2} ln x^{2} - 1 - (- \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣)

Simplify

= \frac{7}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{7}{2} ln x^{2} - 1 + \frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + C

f (x) = 4 x + 1

\int \frac{( x ^{3} + 16 )}{x ^{2} + 16} d x

\int - tan (x) + C d x

\int x \cdot cos (8 x^{2} + π) d x

9 1 + x

What is the integral of (7x-1)/(x^2-1) ?
The integral of (7x-1)/(x^2-1) is 7/2 ln|x^2-1|+1/2 ln|x+1|-1/2 ln|x-1|+C