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

The integral of (7x+1)/(x(x^2-1)) is -3ln|x+1|+4ln|x-1|-ln|x|+C

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

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

- 3 ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣ - ln ∣ x ∣ + C

Solution steps

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

Expand

\frac{7 x + 1}{x ( x ^{2} - 1 )} : \frac{7 x}{x ( x ^{2} - 1 )} + \frac{1}{x ( 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 ( x ^{2} - 1 )} d x + \int \frac{1}{x ( x ^{2} - 1 )} d x

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

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

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

Simplify

- 7 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) - ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣ : - 3 ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣ - ln ∣ x ∣

= - 3 ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣ - ln ∣ x ∣

Add a constant to the solution

= - 3 ln ∣ x + 1 ∣ + 4 ln ∣ x - 1 ∣ - ln ∣ x ∣ + C

\int_{0}^{3} \frac{1}{1 + x} d x

\int \frac{z ^{2}}{1 ^{3}} + z^{3}

\int 20 x^{3} - 12 x^{2} + 6 x d x

n = 1 \sum \infty \frac{7}{n ^{2} + 9}

x \to 0 + lim (\frac{e ^{x}}{e ^{x} - 1})

What is the integral of (7x+1)/(x(x^2-1)) ?
The integral of (7x+1)/(x(x^2-1)) is -3ln|x+1|+4ln|x-1|-ln|x|+C