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)

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

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

= \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

\frac{d}{d x} (2 x (x + 4)^{3})

\frac{\partial}{\partial x} (x^{2} + 3 y^{2} - 6 y + 4 x y)

f (x) = - x^{2} + 2 x + 4, at x = 3

y = - 3 x^{2}

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

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