What is the integral of (2x+1)/((x+1)^3*(x^2+4)^2) ?

The integral of (2x+1)/((x+1)^3*(x^2+4)^2) is

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integral of (2x+1)/((x+1)^3*(x^2+4)^2)

\int \frac{2 x + 1}{( x + 1 ) ^{3} \cdot ( x ^{2} + 4 ) ^{2}} d x

- \frac{31}{10000} arctan (\frac{x}{2}) + \frac{38}{625} ln ∣ x + 1 ∣ - \frac{19}{625} ln x^{2} + 4 - \frac{19}{4000} sin (2 arctan (\frac{x}{2})) - \frac{6}{125 ( x + 1 )} + \frac{1}{50 ( x + 1 ) ^{2}} + \frac{21}{250 ( x ^{2} + 4 )} + C

Solution steps

\int \frac{2 x + 1}{( x + 1 ) ^{3} ( x ^{2} + 4 ) ^{2}} d x

Take the partial fraction of

\frac{2 x + 1}{( x + 1 ) ^{3} ( x ^{2} + 4 ) ^{2}} : \frac{38}{625 ( x + 1 )} + \frac{6}{125 ( x + 1 ) ^{2}} - \frac{1}{25 ( x + 1 ) ^{3}} + \frac{- 38 x + 8}{625 ( x ^{2} + 4 )} + \frac{- 21 x - 19}{125 ( x ^{2} + 4 ) ^{2}}

= \int \frac{38}{625 ( x + 1 )} + \frac{6}{125 ( x + 1 ) ^{2}} - \frac{1}{25 ( x + 1 ) ^{3}} + \frac{- 38 x + 8}{625 ( x ^{2} + 4 )} + \frac{- 21 x - 19}{125 ( x ^{2} + 4 ) ^{2}} 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 \frac{38}{625 ( x + 1 )} d x + \int \frac{6}{125 ( x + 1 ) ^{2}} d x - \int \frac{1}{25 ( x + 1 ) ^{3}} d x + \int \frac{- 38 x + 8}{625 ( x ^{2} + 4 )} d x + \int \frac{- 21 x - 19}{125 ( x ^{2} + 4 ) ^{2}} d x

\int \frac{38}{625 ( x + 1 )} d x = \frac{38}{625} ln ∣ x + 1 ∣

\int \frac{6}{125 ( x + 1 ) ^{2}} d x = - \frac{6}{125 ( x + 1 )}

\int \frac{1}{25 ( x + 1 ) ^{3}} d x = - \frac{1}{50 ( x + 1 ) ^{2}}

\int \frac{- 38 x + 8}{625 ( x ^{2} + 4 )} d x = \frac{1}{625} (- 19 ln x^{2} + 4 + 4 arctan (\frac{x}{2}))

\int \frac{- 21 x - 19}{125 ( x ^{2} + 4 ) ^{2}} d x = \frac{1}{125} (\frac{21}{2 ( x ^{2} + 4 )} - \frac{19}{16} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))))

= \frac{38}{625} ln ∣ x + 1 ∣ - \frac{6}{125 ( x + 1 )} - (- \frac{1}{50 ( x + 1 ) ^{2}}) + \frac{1}{625} (- 19 ln x^{2} + 4 + 4 arctan (\frac{x}{2})) + \frac{1}{125} (\frac{21}{2 ( x ^{2} + 4 )} - \frac{19}{16} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2}))))

Simplify

\frac{38}{625} ln ∣ x + 1 ∣ - \frac{6}{125 ( x + 1 )} - (- \frac{1}{50 ( x + 1 ) ^{2}}) + \frac{1}{625} (- 19 ln x^{2} + 4 + 4 arctan (\frac{x}{2})) + \frac{1}{125} (\frac{21}{2 ( x ^{2} + 4 )} - \frac{19}{16} (arctan (\frac{x}{2}) + \frac{1}{2} sin (2 arctan (\frac{x}{2})))) : - \frac{31}{10000} arctan (\frac{x}{2}) + \frac{38}{625} ln ∣ x + 1 ∣ - \frac{19}{625} ln x^{2} + 4 - \frac{19}{4000} sin (2 arctan (\frac{x}{2})) - \frac{6}{125 ( x + 1 )} + \frac{1}{50 ( x + 1 ) ^{2}} + \frac{21}{250 ( x ^{2} + 4 )}

= - \frac{31}{10000} arctan (\frac{x}{2}) + \frac{38}{625} ln ∣ x + 1 ∣ - \frac{19}{625} ln x^{2} + 4 - \frac{19}{4000} sin (2 arctan (\frac{x}{2})) - \frac{6}{125 ( x + 1 )} + \frac{1}{50 ( x + 1 ) ^{2}} + \frac{21}{250 ( x ^{2} + 4 )}

Add a constant to the solution

= - \frac{31}{10000} arctan (\frac{x}{2}) + \frac{38}{625} ln ∣ x + 1 ∣ - \frac{19}{625} ln x^{2} + 4 - \frac{19}{4000} sin (2 arctan (\frac{x}{2})) - \frac{6}{125 ( x + 1 )} + \frac{1}{50 ( x + 1 ) ^{2}} + \frac{21}{250 ( x ^{2} + 4 )} + C

f (x) = 4^{x}

y = x^{4 x}

\int x tanh^{2} (x) d x

e^{- \frac{1}{x}} + 4 x

f (x) = - \frac{x sin ( \frac{x ^{2}}{4} )}{2}

What is the integral of (2x+1)/((x+1)^3*(x^2+4)^2) ?
The integral of (2x+1)/((x+1)^3*(x^2+4)^2) is