What is the integral of (10)/(x^2-1) ?

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

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integral of (10)/(x^2-1)

\int \frac{10}{x ^{2} - 1} d x

- 10 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

Solution steps

\int \frac{10}{x ^{2} - 1} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 10 \cdot \int \frac{1}{x ^{2} - 1} d x

Factor

x^{2} - 1 : - (- x^{2} + 1)

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

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

= 10 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2}))

Simplify

10 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2})) : - 10 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

= - 10 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣)

Add a constant to the solution

= - 10 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣) + C

y = x^{\frac{1}{2}}, y = x - 2, [0, 4]

\frac{( s + 1 )}{( s ^{2} + 2 s + 2 )}

cos^{3} (t)

\frac{d y}{d x} = \frac{6 y e ^{3 x}}{2 + e ^{3 x}}

\int_{- \infty}^{- 1} e^{- 28 t} d t

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