4x-5-x-2-x-3-x-4-x-5-1-dx-

Question Number 128900 by bemath last updated on 11/Jan/21

\int \frac{4 x + 5}{(x + 2) (x + 3) (x + 4) (x + 5) + 1} dx ?

Answered by liberty last updated on 11/Jan/21

\int \frac{4 x + 5}{(x^{2} + 7 x + 10) (x^{2} + 7 x + 12) + 1} dx = (*)

(x^{2} + 7 x + 10) (x^{2} + 7 x + 12) + 1

let x^{2} + 7 x = u \Rightarrow (u + 10) (u + 12) + 1 = u^{2} + 22 u + 121

= {(u + 11)}^{2} = {(x^{2} + 7 x + 11)}^{2}

(*) \int \frac{4 x + 5}{{(x^{2} + 7 x + 11)}^{2}} dx =

\int \frac{2 (2 x + 7) - 9}{{(x^{2} + 7 x + 11)}^{2}} dx = \int \frac{2 (2 x + 7)}{{(x^{2} + 7 x + 11)}^{2}} dx - \int \frac{9}{{(x^{2} + 7 x + 11)}^{2}} dx

= - \frac{2}{x^{2} + 7 x + 11} - \int \frac{9}{{(x^{2} + 7 x + 11)}^{2}} dx

next . .

Commented by bramlexs22 last updated on 11/Jan/21

I_{2} = \int \frac{9 dx}{{({(x + \frac{7}{2})}^{2} - {(\frac{\sqrt{5}}{2})}^{2})}^{2}}

let x + \frac{7}{2} = \frac{\sqrt{5}}{2} \sec t

I_{2} = \int \frac{\frac{9 \sqrt{5}}{2} \sec t \tan t}{\frac{25}{4} \tan^{4} t} dt

I_{2} = \frac{18 \sqrt{5}}{25} \int \frac{\cos^{2} t}{\sin^{3} t} dt

I_{2} = \frac{18 \sqrt{5}}{25} (\int \csc^{3} tdt - \int \csc t dt)

Answered by Olaf last updated on 11/Jan/21

F (x) = \int \frac{4 x + 5}{(x + 2) (x + 3) (x + 4) (x + 5) + 1} d x

Let u = x + \frac{7}{2}

F (x) = \int \frac{4 u - 9}{(u - \frac{3}{2}) (u - \frac{1}{2}) (u + \frac{1}{2}) (u + \frac{3}{2}) + 1} d u

F (x) = \int \frac{4 u - 9}{(u^{2} - \frac{1}{4}) (u^{2} - \frac{9}{4}) + 1} d u

F (x) = \int \frac{4 u - 9}{u^{4} - \frac{5}{2} u^{2} + \frac{25}{16}} d u

F (x) = \int \frac{4 u - 9}{{(u^{2} - \frac{5}{4})}^{2}} d u

F (x) = \int [\frac{4}{u - \frac{5}{4}} - \frac{4}{{(u^{2} - \frac{5}{4})}^{2}}] d u

F (x) = 4 \ln ∣ u - \frac{5}{4} ∣ + \frac{8 u}{5 (u^{2} - \frac{5}{4})} - \frac{16}{\sqrt{5}} atanh (\frac{2 u}{\sqrt{5}}) + C

F (x) = 4 \ln ∣ x + \frac{9}{4} ∣ + \frac{4 (x + 7)}{5 (x^{2} + 7 x + 11)}

- \frac{16}{\sqrt{5}} atanh (\frac{2 x + 7}{\sqrt{5}}) + C

Answered by MJS_new last updated on 11/Jan/21

\int \frac{4 x + 5}{(x + 2) (x + 3) (x + 4) (x + 5) + 1} d x =

= \int \frac{4 x + 5}{{(x^{2} + 7 x + 11)}^{2}} d x =

[Ostrogradski]

= \frac{18 x + 53}{5 (x^{2} + 7 x + 11)} + \frac{18}{5} \int \frac{d x}{x^{2} + 7 x + 11} =

= \frac{18 x + 53}{5 (x^{2} + 7 x + 11)} + \frac{36 \sqrt{5}}{25} \int (\frac{1}{2 x + 7 - \sqrt{5}} - \frac{1}{2 x + 7 + \sqrt{5}}) d x =

= \frac{18 x + 53}{5 (x^{2} + 7 x + 11)} + \frac{18 \sqrt{5}}{25} (\ln ∣ 2 x + 7 - \sqrt{5} ∣ - \ln ∣ 2 x + 7 + \sqrt{5} ∣ =

= \frac{18 x + 53}{5 (x^{2} + 7 x + 11)} + \frac{18 \sqrt{5}}{25} \ln ∣ \frac{2 x + 7 - \sqrt{5}}{2 x + 7 + \sqrt{5}} ∣ + C

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