What is the integral of (x^3+6x^2+2x+3)/(x^2+6x+1) ?

The integral of (x^3+6x^2+2x+3)/(x^2+6x+1) is

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integral of (x^3+6x^2+2x+3)/(x^2+6x+1)

\int \frac{x ^{3} + 6 x ^{2} + 2 x + 3}{x ^{2} + 6 x + 1} d x

\frac{x ^{2}}{2} + \frac{2 2 + 3}{4 2} ln x + 3 + 22 + \frac{2 2 - 3}{4 2} ln x + 3 - 22 - \frac{3}{4 2} (ln \frac{x + 3}{2 2} + 1 - ln \frac{x + 3}{2 2} - 1) + C

Solution steps

\int \frac{x ^{3} + 6 x ^{2} + 2 x + 3}{x ^{2} + 6 x + 1} d x

Long division

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

= \int x + \frac{x + 3}{x ^{2} + 6 x + 1} d x

Apply the Sum Rule:

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

= \int x d x + \int \frac{x + 3}{x ^{2} + 6 x + 1} d x

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

\int \frac{x + 3}{x ^{2} + 6 x + 1} d x = \frac{2 2 + 3}{4 2} ln x + 3 + 22 + \frac{2 2 - 3}{4 2} ln x + 3 - 22 - \frac{3}{4 2} (ln \frac{x + 3}{2 2} + 1 - ln \frac{x + 3}{2 2} - 1)

= \frac{x ^{2}}{2} + \frac{2 2 + 3}{4 2} ln x + 3 + 22 + \frac{2 2 - 3}{4 2} ln x + 3 - 22 - \frac{3}{4 2} (ln \frac{x + 3}{2 2} + 1 - ln \frac{x + 3}{2 2} - 1)

Add a constant to the solution

= \frac{x ^{2}}{2} + \frac{2 2 + 3}{4 2} ln x + 3 + 22 + \frac{2 2 - 3}{4 2} ln x + 3 - 22 - \frac{3}{4 2} (ln \frac{x + 3}{2 2} + 1 - ln \frac{x + 3}{2 2} - 1) + C

\int \frac{1}{49 e ^{- 7 x} + e ^{7 x}} d x

\frac{d y}{d x} = e^{x^{2}}

(ln (x))^{4}

2 x y \frac{d y}{d x} + y^{2} = 2 x^{2}

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