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

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

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

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

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

Solution steps

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

Complete the square

x^{2} + 2 x + 1 : (x + 1)^{2}

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

Apply u-substitution

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

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

Expand

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

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

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

\int 3 d u = 3 u

\int \frac{3}{u} d u = 3 ln ∣ u ∣

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

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

Substitute back

u = x + 1

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

Simplify

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

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

Add a constant to the solution

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

y = \frac{( x ^{2} )}{8 + x}

\int a e^{- m x} d x

\frac{d y}{d x} + (\frac{3}{x}) y = - 2 x + 7

x \to - 1 lim (1 + 2 x)

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