x-2-dx-x-1-x-2-4-2-

Question Number 7422 by Tawakalitu. last updated on 28/Aug/16

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

Answered by Yozzia last updated on 28/Aug/16

\frac{x^{2}}{(x - 1) {(x^{2} + 4)}^{2}} \equiv \frac{a}{x - 1} + \frac{b x + c}{x^{2} + 4} + \frac{e x + f}{{(x^{2} + 4)}^{2}}

\frac{x^{2}}{(x - 1) {(x^{2} + 4)}^{2}} \equiv \frac{a {(x^{2} + 4)}^{2} + (b x + c) (x - 1) (x^{2} + 4) + (e x + f) (x - 1)}{(x - 1) {(x^{2} + 4)}^{2}}

∴ x^{2} = a {(x^{2} + 4)}^{2} + (b x + c) (x - 1) (x^{2} + 4) + (e x + f) (x - 1)

x = 1 \Rightarrow 1 = a {(5)}^{2} + 0 \Rightarrow a = \frac{1}{25}

x = 2 i \Rightarrow - 4 = a {(- 4 + 4)}^{2} + 0 + (2 e i + f) (2 i - 1)

∴ f + 2 e i = \frac{- 4}{2 i - 1} = \frac{- 4 (- 2 i - 1)}{5}

x = - 2 i \Rightarrow - 4 = 0 + 0 + (- 2 i e + f) (- 2 i - 1)

∴ f - 2 e i = \frac{4}{1 + 2 i} = \frac{4 (1 - 2 i)}{5}

∴ 2 f + 0 = \frac{4 (1 - 2 i + 2 i + 1)}{5} = \frac{8}{5} \Rightarrow f = \frac{4}{5}

∴ \frac{4}{5} + 2 e i = \frac{4}{5} (1 + 2 i) \Rightarrow e = \frac{4}{5}

x = 0 \Rightarrow 0 = \frac{16}{25} + c (- 1) (4) + \frac{4}{5} (0 + 1) (- 1)

0 = \frac{16}{25} - \frac{4}{5} - 4 c

\Rightarrow c = \frac{4}{25} - \frac{1}{5} = \frac{20 - 25}{125} = \frac{- 1}{25}

x = 2 \Rightarrow 4 = \frac{1}{25} 64 + (2 b - \frac{1}{25}) (1) (8) + \frac{4}{5} (2 + 1) (1)

4 = \frac{64}{25} + 16 b - \frac{8}{25} + \frac{60}{25} = \frac{116}{25} + 16 b

b = \frac{1}{16} (4 - \frac{116}{25}) = \frac{- 1}{25}

∴ \frac{x^{2}}{(x - 1) {(x^{2} + 4)}^{2}} = \frac{1}{25 (x - 1)} - \frac{x + 1}{25 (x^{2} + 4)} + \frac{4 (x + 1)}{5 {(x^{2} + 4)}^{2}}

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

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

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(1) \int \frac{1}{25 (x - 1)} d x = \frac{1}{25} l n ∣ x - 1 ∣ + c (1)

(2) \frac{1}{25} \int \frac{2 x}{x^{2} + 4} d x = \frac{1}{25} l n (x^{2} + 4) + c (2)

(3) \frac{1}{25} \int \frac{1}{x^{2} + 4} d x = \frac{1}{50} {t a n}^{- 1} \frac{x}{2} + c (3)

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

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

(5) \int \frac{4}{5 {(x^{2} + 4)}^{2}} d x [x = 2 t a n k \Rightarrow d x = 2 {s e c}^{2} k d k]

\int \frac{4}{5 {(x^{2} + 4)}^{2}} d x = \int \frac{8 {s e c}^{2} k d k}{5 {(4 {t a n}^{2} k + 4)}^{2}} = \frac{8}{80} \int \frac{{s e c}^{2} k d k}{{s e c}^{4} k} = \frac{1}{10} \int {c o s}^{2} k d k = \frac{1}{20} \int (1 + c o s 2 k) d k

= \frac{1}{20} (k + \frac{1}{2} s i n 2 k)

\int \frac{4}{5 {(x^{2} + 4)}^{2}} d x = \frac{1}{20} k + \frac{1}{20} s i n k c o s k

∵ t a n k = \frac{x}{2} \Rightarrow s i n k = \frac{x}{\sqrt{x^{2} + 4}}, c o s k = \frac{2}{\sqrt{x^{2} + 4}}

\int \frac{4}{5 {(x^{2} + 4)}^{2}} d x = \frac{1}{20} {t a n}^{- 1} \frac{x}{2} + \frac{x}{10 (x^{2} + 4)} + c (5)

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∴ \int \frac{x^{2}}{(x - 1) {(x^{2} + 4)}^{2}} d x = \frac{1}{25} l n ∣ x - 1 ∣ - \frac{1}{50} l n (x^{2} + 4) + \frac{50 - 20}{1000} {t a n}^{- 1} \frac{x}{2} - \frac{2}{5 (x^{2} + 4)} + \frac{x}{10 (x^{2} + 4)} + C

∴ \int \frac{x^{2}}{(x - 1) {(x^{2} + 4)}^{2}} d x = \frac{1}{25} l n ∣ x - 1 ∣ - \frac{1}{50} l n (x^{2} + 4) + \frac{3}{100} {t a n}^{- 1} \frac{x}{2} + \frac{x - 4}{10 (x^{2} + 4)} + C

Commented by Tawakalitu. last updated on 28/Aug/16

W o w, t h a n k s s o m u c h s i r . i r e a l l y a p p r e c i a t e

Commented by Tawakalitu. last updated on 28/Aug/16

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