What is the integral of (x^3-x-1)/((x+1)^2) ?

The integral of (x^3-x-1)/((x+1)^2) is -3x+2ln|x+1|+1/2 (x+1)^2+1/(x+1)-3+C

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integral of (x^3-x-1)/((x+1)^2)

\int \frac{x ^{3} - x - 1}{( x + 1 ) ^{2}} d x

- 3 x + 2 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} + \frac{1}{x + 1} - 3 + C

Solution steps

\int \frac{x ^{3} - x - 1}{( x + 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{- u + ( u - 1 ) ^{3}}{u ^{2}} d u

= \int \frac{- u + ( u - 1 ) ^{3}}{u ^{2}} d u

Expand

\frac{- u + ( u - 1 ) ^{3}}{u ^{2}} : - \frac{1}{u} + \frac{( u - 1 ) ^{3}}{u ^{2}}

= \int - \frac{1}{u} + \frac{( u - 1 ) ^{3}}{u ^{2}} 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 \frac{1}{u} d u + \int \frac{( u - 1 ) ^{3}}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{( u - 1 ) ^{3}}{u ^{2}} d u = \frac{u ^{2}}{2} - 3 u + 3 ln ∣ u ∣ + \frac{1}{u}

= - ln ∣ u ∣ + \frac{u ^{2}}{2} - 3 u + 3 ln ∣ u ∣ + \frac{1}{u}

Substitute back

u = x + 1

= - ln ∣ x + 1 ∣ + \frac{( x + 1 ) ^{2}}{2} - 3 (x + 1) + 3 ln ∣ x + 1 ∣ + \frac{1}{x + 1}

Simplify

- ln ∣ x + 1 ∣ + \frac{( x + 1 ) ^{2}}{2} - 3 (x + 1) + 3 ln ∣ x + 1 ∣ + \frac{1}{x + 1} : - 3 x + 2 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} + \frac{1}{x + 1} - 3

= - 3 x + 2 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} + \frac{1}{x + 1} - 3

Add a constant to the solution

= - 3 x + 2 ln ∣ x + 1 ∣ + \frac{1}{2} (x + 1)^{2} + \frac{1}{x + 1} - 3 + C

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\int (x + \frac{sin ( x )}{x}) d x

cos (π x)

\int 5 x^{2} cos (7 x) d x

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What is the integral of (x^3-x-1)/((x+1)^2) ?
The integral of (x^3-x-1)/((x+1)^2) is -3x+2ln|x+1|+1/2 (x+1)^2+1/(x+1)-3+C