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

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

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

integral

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

Apply the Sum Rule:

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

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

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

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

\int 3 d v = 3 v

\int \frac{1}{v} d v = ln ∣ v ∣

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

Substitute back

v = u + 1

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

Add a constant to the solution

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

(cos (x))^{2}

1 - y^{2}

e^{x + 1}

x^{5} (x^{2} + 3)^{4}

- 3 x^{- 4}

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