What is the integral of (3x^3+6x^2+3)/(3x+3) ?

The integral of (3x^3+6x^2+3)/(3x+3) is x^2+x+2ln|x+1|+1/3 (x+1)^3-3/2 (x+1)^2+C

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integral of (3x^3+6x^2+3)/(3x+3)

\int \frac{3 x ^{3} + 6 x ^{2} + 3}{3 x + 3} d x

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

Solution steps

\int \frac{3 x ^{3} + 6 x ^{2} + 3}{3 x + 3} d x

Simplify

\frac{3 x ^{3} + 6 x ^{2} + 3}{3 x + 3} : \frac{x ^{3} + 2 x ^{2} + 1}{x + 1}

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

Expand

\frac{x ^{3} + 2 x ^{2} + 1}{x + 1} : \frac{x ^{3}}{x + 1} + \frac{2 x ^{2}}{x + 1} + \frac{1}{x + 1}

Apply the Sum Rule:

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

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

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

\int \frac{2 x ^{2}}{x + 1} d x = x^{2} - 2 x + 2 ln ∣ x + 1 ∣ - 3

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

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

Simplify

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

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

Add a constant to the solution

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

x \to \infty lim (0.6 6^{x})

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

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + y})

\frac{s}{4 s ^{2} + 1}

What is the integral of (3x^3+6x^2+3)/(3x+3) ?
The integral of (3x^3+6x^2+3)/(3x+3) is x^2+x+2ln|x+1|+1/3 (x+1)^3-3/2 (x+1)^2+C