What is the integral of (6x+6)/(5x^2+26x+5) ?

The integral of (6x+6)/(5x^2+26x+5) is

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integral of (6x+6)/(5x^2+26x+5)

\int \frac{6 x + 6}{5 x ^{2} + 26 x + 5} d x

6 (\frac{1}{10} ln 25 x^{2} + 130 x + 25 + \frac{13}{120} (ln \frac{25}{12} + \frac{5}{12} x - ln \frac{1}{12} + \frac{5}{12} x)) - \frac{1}{4} (ln \frac{25}{12} + \frac{5}{12} x - ln \frac{1}{12} + \frac{5}{12} x) + C

Solution steps

\int \frac{6 x + 6}{5 x ^{2} + 26 x + 5} d x

Expand

\frac{6 x + 6}{5 x ^{2} + 26 x + 5} : \frac{6 x}{5 x ^{2} + 26 x + 5} + \frac{6}{5 x ^{2} + 26 x + 5}

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{6 x}{5 x ^{2} + 26 x + 5} d x + \int \frac{6}{5 x ^{2} + 26 x + 5} d x

\int \frac{6 x}{5 x ^{2} + 26 x + 5} d x = 6 (\frac{1}{10} ln 25 x^{2} + 130 x + 25 + \frac{13}{120} (ln \frac{25}{12} + \frac{5}{12} x - ln \frac{1}{12} + \frac{5}{12} x))

\int \frac{6}{5 x ^{2} + 26 x + 5} d x = - \frac{1}{4} (ln \frac{25}{12} + \frac{5}{12} x - ln \frac{1}{12} + \frac{5}{12} x)

= 6 (\frac{1}{10} ln 25 x^{2} + 130 x + 25 + \frac{13}{120} (ln \frac{25}{12} + \frac{5}{12} x - ln \frac{1}{12} + \frac{5}{12} x)) - \frac{1}{4} (ln \frac{25}{12} + \frac{5}{12} x - ln \frac{1}{12} + \frac{5}{12} x)

Add a constant to the solution

= 6 (\frac{1}{10} ln 25 x^{2} + 130 x + 25 + \frac{13}{120} (ln \frac{25}{12} + \frac{5}{12} x - ln \frac{1}{12} + \frac{5}{12} x)) - \frac{1}{4} (ln \frac{25}{12} + \frac{5}{12} x - ln \frac{1}{12} + \frac{5}{12} x) + C

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

\frac{\partial}{\partial x} (sin (x^{3} + y^{2}))

y^{^{'}} + 5 y = 1

(- 3, 2), (5, - 4)

x \to - 1 lim (x^{2} + x + 1)

What is the integral of (6x+6)/(5x^2+26x+5) ?
The integral of (6x+6)/(5x^2+26x+5) is