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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{( u - 2 ) ^{2}}{u} d u

= \int \frac{( u - 2 ) ^{2}}{u} d u

Expand

\frac{( u - 2 ) ^{2}}{u} : u - 4 + \frac{4}{u}

= \int u - 4 + \frac{4}{u} d u

Apply the Sum Rule:

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

= \int u d u - \int 4 d u + \int \frac{4}{u} d u

\int u d u = \frac{u ^{2}}{2}

\int 4 d u = 4 u

\int \frac{4}{u} d u = 4 ln ∣ u ∣

= \frac{u ^{2}}{2} - 4 u + 4 ln ∣ u ∣

Substitute back

u = x + 1

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

Add a constant to the solution

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

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