What is the integral of (2x^3)/(x+1) ?

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

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integral of (2x^3)/(x+1)

\int \frac{2 x ^{3}}{x + 1} d x

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

Solution steps

\int \frac{2 x ^{3}}{x + 1} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{x ^{3}}{x + 1} d x

Apply u-substitution

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

Expand

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

= 2 \cdot \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

= 2 (\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 ∣

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

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