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

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

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

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

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

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

(0, 2), (- 1, - 1)

\int_{- 1}^{1} (t^{2} - 2) d t

x \to \infty lim (- 5 π x)

y = \frac{1}{x - 8}, (- 1, - \frac{1}{9})

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

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