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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Factor

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

= 4 \cdot \int \frac{x ^{2}}{- ( - x ^{2} + 1 )} d x

Take the constant out:

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

= 4 (- \int \frac{x ^{2}}{- x ^{2} + 1} d x)

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

= 4 (- \int \frac{1}{- x ^{2} + 1} - 1 d x)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (- (\int \frac{1}{- x ^{2} + 1} d x - \int 1 d x))

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

\int 1 d x = x

= 4 (- (\frac{ln ∣ x + 1 ∣}{2} - \frac{ln ∣ x - 1 ∣}{2} - x))

Simplify

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

= - 4 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x)

Add a constant to the solution

= - 4 (\frac{1}{2} ln ∣ x + 1 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ - x) + C

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