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

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

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

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

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

Solution steps

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

Multiply by the conjugate of

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

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

Multiply

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

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

Factor

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

= \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

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

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

= - \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

= - (\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

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

Simplify

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

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

Add a constant to the solution

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

\int 12 x z cos (x z^{2}) d z

f (x) = ln (1 + sin^{2} (x))

x \to - 0 lim (x^{2})

\int e^{x} (x^{2} - 2 x - 1) d x

\frac{\partial}{\partial x} (x ln (2 y - x))

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