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

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

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

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

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

Solution steps

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

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 = 1 + x

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

Add a constant to the solution

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

y^{^{''}} - y^{^{'}} - 2 y = - 6 t + 4 t^{2}

y = sin (x), (4 π, 0)

\frac{d y}{d x} = x y, y (0) = 0

f (x) = 3 x, (1, 3)

\int_{- 2}^{4} ∣ - 2 x + 4 ∣ d x

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