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

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

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

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

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

Solution steps

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

Simplify

x^{3} \frac{1}{x + 1} : \frac{x ^{3}}{x + 1}

= \int \frac{x ^{3}}{x + 1} d x

Apply u-substitution

= \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 = x + 1

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

Add a constant to the solution

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

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