What is the integral of (5x+5)/(4x^2+17x+4) ?

The integral of (5x+5)/(4x^2+17x+4) is

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integral of (5x+5)/(4x^2+17x+4)

\int \frac{5 x + 5}{4 x ^{2} + 17 x + 4} d x

10 (\frac{1}{16} ln 64 x^{2} + 272 x + 64 + \frac{17}{240} (ln \frac{32}{15} + \frac{8}{15} x - ln \frac{2}{15} + \frac{8}{15} x)) - \frac{1}{3} (ln \frac{32}{15} + \frac{8}{15} x - ln \frac{2}{15} + \frac{8}{15} x) + C

Solution steps

\int \frac{5 x + 5}{4 x ^{2} + 17 x + 4} d x

Expand

\frac{5 x + 5}{4 x ^{2} + 17 x + 4} : \frac{5 x}{4 x ^{2} + 17 x + 4} + \frac{5}{4 x ^{2} + 17 x + 4}

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{5 x}{4 x ^{2} + 17 x + 4} d x + \int \frac{5}{4 x ^{2} + 17 x + 4} d x

\int \frac{5 x}{4 x ^{2} + 17 x + 4} d x = 10 (\frac{1}{16} ln 64 x^{2} + 272 x + 64 + \frac{17}{240} (ln \frac{32}{15} + \frac{8}{15} x - ln \frac{2}{15} + \frac{8}{15} x))

\int \frac{5}{4 x ^{2} + 17 x + 4} d x = - \frac{1}{3} (ln \frac{32}{15} + \frac{8}{15} x - ln \frac{2}{15} + \frac{8}{15} x)

= 10 (\frac{1}{16} ln 64 x^{2} + 272 x + 64 + \frac{17}{240} (ln \frac{32}{15} + \frac{8}{15} x - ln \frac{2}{15} + \frac{8}{15} x)) - \frac{1}{3} (ln \frac{32}{15} + \frac{8}{15} x - ln \frac{2}{15} + \frac{8}{15} x)

Add a constant to the solution

= 10 (\frac{1}{16} ln 64 x^{2} + 272 x + 64 + \frac{17}{240} (ln \frac{32}{15} + \frac{8}{15} x - ln \frac{2}{15} + \frac{8}{15} x)) - \frac{1}{3} (ln \frac{32}{15} + \frac{8}{15} x - ln \frac{2}{15} + \frac{8}{15} x) + C

\frac{d}{d x} (cos (x) sin (y))

\frac{1}{x ^{2}}, 4

y^{^{'}} = y (x y^{6} + 5)

\int_{1}^{9} 2^{x} d x

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

What is the integral of (5x+5)/(4x^2+17x+4) ?
The integral of (5x+5)/(4x^2+17x+4) is