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

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

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

integral

\frac{x ^{3}}{( x + 1 ) ^{3}}

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

Solution steps

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

Apply u-substitution

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

Expand

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

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

\int 1 d u = u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

\int \frac{3}{u ^{2}} d u = - \frac{3}{u}

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

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

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

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