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

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

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

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

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

Solution steps

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

Expand

\frac{2 x + 1}{x ^{2} - 1} : \frac{2 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{2 x}{x ^{2} - 1} d x + \int \frac{1}{x ^{2} - 1} d x

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

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

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

Add a constant to the solution

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

\int csc^{3} (d) x d x

\int_{0}^{1} x^{4} e^{x} d x

\int \frac{26}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{\partial}{\partial y} (sin (x y^{2} + y) + 2 e^{x})

n = 1 \sum \infty \frac{1 + 5 ^{n}}{7 ^{n}}

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