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

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

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

\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) + C

Solution steps

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

Expand

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

\int \frac{x}{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

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

= \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{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 y (2 y^{2} - 3) d y

\frac{d}{d x} (4^{- 2 x})

\frac{d ^{2}}{d x ^{2}} y + 81 y = 0

\frac{d}{d x} (\frac{1 - cosh ( x )}{1 + cosh ( x )})

\int_{0}^{1} x^{2} (x^{3} + 1)^{3} d x