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

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

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

\int \frac{w ^{3}}{w + 1} d w

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

Solution steps

\int \frac{w ^{3}}{w + 1} d w

Apply u-substitution

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

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

Expand

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

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

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

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

\int 3 d u = 3 u

\int \frac{1}{u} d u = ln ∣ u ∣

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

Substitute back

u = w + 1

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

Add a constant to the solution

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

\frac{d}{d x} (19 arctan (x))

ln (\frac{6 x ^{2} + 5}{x ^{\frac{4}{5}}})

\frac{s + 4}{s ^{2} + 2 s - 3}

5 x^{2} + 2 x + 6

\frac{\partial}{\partial x} (\frac{x ^{2} + y ^{2}}{2})

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