What is the integral of (2x^2+4x-5)/(x(x^2+6x+10)^2) ?

The integral of (2x^2+4x-5)/(x(x^2+6x+10)^2) is

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integral of (2x^2+4x-5)/(x(x^2+6x+10)^2)

\int \frac{2 x ^{2} + 4 x - 5}{x ( x ^{2} + 6 x + 10 ) ^{2}} d x

\frac{5}{8} x sin (2 arctan (x + 3)) + \frac{7}{4} sin (2 arctan (x + 3)) - \frac{1}{10} arctan (x + 3) - \frac{1}{20} ln ∣ x ∣ + \frac{1}{40} ln x^{2} + 6 x + 10 + C

Solution steps

\int \frac{2 x ^{2} + 4 x - 5}{x ( x ^{2} + 6 x + 10 ) ^{2}} d x

Expand

\frac{2 x ^{2} + 4 x - 5}{x ( x ^{2} + 6 x + 10 ) ^{2}} : \frac{2 x ^{2}}{x ( x ^{2} + 6 x + 10 ) ^{2}} + \frac{4 x}{x ( x ^{2} + 6 x + 10 ) ^{2}} - \frac{5}{x ( x ^{2} + 6 x + 10 ) ^{2}}

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{2 x ^{2}}{x ( x ^{2} + 6 x + 10 ) ^{2}} d x + \int \frac{4 x}{x ( x ^{2} + 6 x + 10 ) ^{2}} d x - \int \frac{5}{x ( x ^{2} + 6 x + 10 ) ^{2}} d x

\int \frac{2 x ^{2}}{x ( x ^{2} + 6 x + 10 ) ^{2}} d x = \frac{1}{2} x sin (2 arctan (x + 3)) - 3 arctan (x + 3)

\int \frac{4 x}{x ( x ^{2} + 6 x + 10 ) ^{2}} d x = 2 arctan (x + 3) + sin (2 arctan (x + 3))

\int \frac{5}{x ( x ^{2} + 6 x + 10 ) ^{2}} d x = 5 (- \frac{1}{40} x sin (2 arctan (x + 3)) - \frac{9}{50} arctan (x + 3) - \frac{3}{20} sin (2 arctan (x + 3)) + \frac{1}{100} ln ∣ x ∣ - \frac{1}{200} ln x^{2} + 6 x + 10)

= \frac{1}{2} x sin (2 arctan (x + 3)) - 3 arctan (x + 3) + 2 arctan (x + 3) + sin (2 arctan (x + 3)) - 5 (- \frac{1}{40} x sin (2 arctan (x + 3)) - \frac{9}{50} arctan (x + 3) - \frac{3}{20} sin (2 arctan (x + 3)) + \frac{1}{100} ln ∣ x ∣ - \frac{1}{200} ln x^{2} + 6 x + 10)

Simplify

\frac{1}{2} x sin (2 arctan (x + 3)) - 3 arctan (x + 3) + 2 arctan (x + 3) + sin (2 arctan (x + 3)) - 5 (- \frac{1}{40} x sin (2 arctan (x + 3)) - \frac{9}{50} arctan (x + 3) - \frac{3}{20} sin (2 arctan (x + 3)) + \frac{1}{100} ln ∣ x ∣ - \frac{1}{200} ln x^{2} + 6 x + 10) : \frac{5}{8} x sin (2 arctan (x + 3)) + \frac{7}{4} sin (2 arctan (x + 3)) - \frac{1}{10} arctan (x + 3) - \frac{1}{20} ln ∣ x ∣ + \frac{1}{40} ln x^{2} + 6 x + 10

= \frac{5}{8} x sin (2 arctan (x + 3)) + \frac{7}{4} sin (2 arctan (x + 3)) - \frac{1}{10} arctan (x + 3) - \frac{1}{20} ln ∣ x ∣ + \frac{1}{40} ln x^{2} + 6 x + 10

Add a constant to the solution

= \frac{5}{8} x sin (2 arctan (x + 3)) + \frac{7}{4} sin (2 arctan (x + 3)) - \frac{1}{10} arctan (x + 3) - \frac{1}{20} ln ∣ x ∣ + \frac{1}{40} ln x^{2} + 6 x + 10 + C

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What is the integral of (2x^2+4x-5)/(x(x^2+6x+10)^2) ?
The integral of (2x^2+4x-5)/(x(x^2+6x+10)^2) is