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Question Number 118997 by Lordose last updated on 21/Oct/20

	Answered by MJS_new last updated on 21/Oct/20

	$\int \frac{x^{2} - 2}{{(x^{2} + 2)}^{5}} d x =$ $[{Ostrogradski}^{'} s Method]$ $= - \frac{x (15 x^{6} + 110 x^{4} + 292 x^{2} + 392)}{512 {(x^{2} + 2)}^{4}} - \frac{15}{512} \int \frac{d x}{x^{2} + 2} =$ $= - \frac{x (15 x^{6} + 110 x^{4} + 292 x^{2} + 392)}{512 {(x^{2} + 2)}^{4}} - \frac{15 \sqrt{2}}{1024} \arctan \frac{\sqrt{2} x}{2} + C$
	Commented by MJS_new last updated on 21/Oct/20

	$I used and explained {Ostrogradski}^{'} s method$ $several times before, search for ‘ ‘ {Ostrogradski}^{″}$ $here on the forum, I think {you}^{'} ll find enough$ $info there$
	Commented by Ar Brandon last updated on 21/Oct/20
	Sir it seems you deleted it before the arrival of "Her Majesty"�� alongside some other posts. Greetings Sir��
	Commented by Lordose last updated on 22/Oct/20
	Thanks sire. I also used ostrogradski method to resolve the integral.
	Commented by Lordose last updated on 22/Oct/20
	Nope. It's still in the forum. I thought you speak French ��
	Commented by Ar Brandon last updated on 22/Oct/20
	Are you referring to me about speaking French, Lordose ?
	Commented by Ar Brandon last updated on 22/Oct/20
	I do speak French. By-the-way it's the most spoken language in my country��
	Commented by MJS_new last updated on 22/Oct/20
	I'm from Austria, my native language is German. I have been speaking English for almost 5 decades now. I learned French in school for 4 years but I never used it and forgot most of it. Reading it I'm able to get the meaning somehow...
	Commented by Dwaipayan Shikari last updated on 22/Oct/20
	Mozart and Beethoven were Austrian and German
	Answered by 1549442205PVT last updated on 22/Oct/20
	$I_n =∫(dx/((x^2 +a^2 )^n ))=(1/a^2 )∫((a^2 +x^2 −x^2 dx)/((x^2 +a^2 )^n )) =(1/a^2 )I_(n−1) −(1/a^2 )∫((x.xdx)/((x^2 +a^2 )^5 )) .Put u=x,dv=((xdx)/([x^2 +a^2 ]^n )),du=dx ⇒v=(1/(2a^2 ))∫((d(x^2 +a^2 ))/([x^2 +a^2 ]^n ))= −(1/(2a^2 (n−1))).(1/((x^2 +a^2 )^(n−1) )).Hence I_n =(1/a^2 )I_(n−1) +(1/a^2 )[(x/(2a^2 (n−1)(x^2 +a^2 )^(n−1) ))− (1/(2(n−1)))∫(dx/((x^2 +a^2 )^(n−1) ))] or I_n =(1/a^2 )I_(n−1) +(x/(2a^2 (n−1)(x^2 +a^2 )^(n−1) ))−(1/(2a^2 (n−1)))I_(n−1) I_n =(x/(2a^2 (n−1)(x^2 +a^2 )^(n−1) ))+(1/(2a^2 )).((2n−3)/(2n−2))I_(n−1) Apply on the above problem we have F=∫((x^2 −2)/((x^2 +2)^5 ))dx=∫(dx/((x^2 +2)^4 ))−4∫(dx/((x^2 +2)^5 )) F=I_4 −4I_5 I_5 =(x/(4.4(x^2 +2)^4 ))+(1/4).(7/8).I_4 ⇒F=I_4 −4I_5 =I_4 −(x/(4(x^2 +2)^4 ))−(7/8)I_4 =−(x/(4(x^2 +2)^4 ))+(1/8)I_4 I_4 =(x/(4.3(x^2 +2)^3 ))+(1/4).(5/6){(x/(4.2(x^2 +2)^2 ))+ (1/4).(3/4)[(x/(4.2(x^2 +2)))+(1/4).(1/2).(1/( (√2)))tan^(−1) (x/( (√2)))]} =(x/(12(x^2 +2)^3 ))+((5x)/(192(x^2 +2)^2 ))+((15x)/(3072(x^2 +2))) +((15)/(3072(√2)))tan^(−1) (x/( (√2))) F=((−x)/(4(x^2 +2)^4 ))+(x/(96(x^2 +2)^3 ))+((5x)/(1536(x^2 +2)^2 ))+ +((15x)/(24576(x^2 +2)))+((15)/( 24576(√2)))tan^(−1) (x/( (√2)))+C$
	$I_{n} = \int \frac{dx}{{(x^{2} + a^{2})}^{n}} = \frac{1}{a^{2}} \int \frac{a^{2} + x^{2} - x^{2} dx}{{(x^{2} + a^{2})}^{n}}$ $= \frac{1}{a^{2}} I_{n - 1} - \frac{1}{a^{2}} \int \frac{x . xdx}{{(x^{2} + a^{2})}^{5}}$ $. Put u = x, dv = \frac{xdx}{{[x^{2} + a^{2}]}^{n}}, du = dx$ $\Rightarrow v = \frac{1}{{2 a}^{2}} \int \frac{d (x^{2} + a^{2})}{{[x^{2} + a^{2}]}^{n}} =$ $- \frac{1}{{2 a}^{2} (n - 1)} . \frac{1}{{(x^{2} + a^{2})}^{n - 1}} . Hence$ $I_{n} = \frac{1}{a^{2}} I_{n - 1} + \frac{1}{a^{2}} [\frac{x}{{2 a}^{2} (n - 1) {(x^{2} + a^{2})}^{n - 1}} -$ $\frac{1}{2 (n - 1)} \int \frac{dx}{{(x^{2} + a^{2})}^{n - 1}}] or$ $I_{n} = \frac{1}{a^{2}} I_{n - 1} + \frac{x}{{2 a}^{2} (n - 1) {(x^{2} + a^{2})}^{n - 1}} - \frac{1}{{2 a}^{2} (n - 1)} I_{n - 1}$ $I_{n} = \frac{x}{2 a^{2} (n - 1) {(x^{2} + a^{2})}^{n - 1}} + \frac{1}{2 a^{2}} . \frac{2 n - 3}{2 n - 2} I_{n - 1}$ $Apply on the above problem we have$ $F = \int \frac{x^{2} - 2}{{(x^{2} + 2)}^{5}} dx = \int \frac{dx}{{(x^{2} + 2)}^{4}} - 4 \int \frac{dx}{{(x^{2} + 2)}^{5}}$ $F = I_{4} - {4 I}_{5}$ $I_{5} = \frac{x}{4.4 {(x^{2} + 2)}^{4}} + \frac{1}{4} . \frac{7}{8} . I_{4}$ $\Rightarrow F = I_{4} - {4 I}_{5} = I_{4} - \frac{x}{4 {(x^{2} + 2)}^{4}} - \frac{7}{8} I_{4}$ $= - \frac{x}{4 {(x^{2} + 2)}^{4}} + \frac{1}{8} I_{4}$ $I_{4} = \frac{x}{4.3 {(x^{2} + 2)}^{3}} + \frac{1}{4} . \frac{5}{6} {\frac{x}{4.2 {(x^{2} + 2)}^{2}} +$ $\frac{1}{4} . \frac{3}{4} [\frac{x}{4.2 (x^{2} + 2)} + \frac{1}{4} . \frac{1}{2} . \frac{1}{\sqrt{2}} \tan^{- 1} \frac{x}{\sqrt{2}}]}$ $= \frac{x}{12 {(x^{2} + 2)}^{3}} + \frac{5 x}{192 {(x^{2} + 2)}^{2}} + \frac{15 x}{3072 (x^{2} + 2)}$ $+ \frac{15}{3072 \sqrt{2}} \tan^{- 1} \frac{x}{\sqrt{2}}$ $F = \frac{- x}{4 {(x^{2} + 2)}^{4}} + \frac{x}{96 {(x^{2} + 2)}^{3}} + \frac{5 x}{1536 {(x^{2} + 2)}^{2}} +$ $+ \frac{15 x}{24576 (x^{2} + 2)} + \frac{15}{24576 \sqrt{2}} \tan^{- 1} \frac{x}{\sqrt{2}} + C$
	Answered by Bird last updated on 22/Oct/20

	$I = \int \frac{x^{2} - 2}{{(x^{2} + 2)}^{5}} d x \Rightarrow$ $I ? = \int \frac{x^{2} - 2}{{(x - i \sqrt{2})}^{5} {(x + i \sqrt{2})}^{5}} d x$ $= \int \frac{x^{2} - 2}{{(\frac{x - i \sqrt{2}}{x + i \sqrt{2}})}^{5} {(x + i \sqrt{2})}^{10}} d x w e d o$ $t h e c h a n g . \frac{x - i \sqrt{2}}{x + i \sqrt{2}} = t \Rightarrow x - i \sqrt{2} = t x + i t \sqrt{2}$
	Commented by Bird last updated on 22/Oct/20
	$⇒(1−t)x=(1+t)i(√2) ⇒x=i(√2)×((1+t)/(1−t)) (dx/dt)=i(√2)×((1−t+(1+t))/((1−t)^2 )) =((2i(√2))/((1−t)^2 )) x+i(√2)=i(√(2())((1+t)/(1−t))+1) =i(√2)×(2/(1−t)) =((2i(√2))/(1−t)) ⇒ I =∫ ((−2(((1+t)/(1−t)))^2 −2)/(t^5 (2i(√2))^(10) ((1/(1−t)))^(10) ))×((2i(√2))/((1−t)^2 )) =−2(1/((2i(√2))^9 ))∫ (((((1+t)/(1−t)))^2 +1)/(t^5 (1−t)^2 ))×(1−t)^(10) dt =−(2/((2i(√2))^9 )) ∫ (((1−t)^8 { (((1+t)^2 )/((1−t)^2 ))+1})/t^5 )dt =−(2/((2i(√2))^9 ))∫ (((1−t)^8 ((1+t)^2 +(1−t)^2 ))/(t^5 (1−t)^2 ))dt =((−2)/((2i(√2))^9 ))∫(((1−t)^6 ( t^(2 ) +2t+1+t^2 −2t+1))/t^5 )dt =((−2)/((2i(√2))^9 ))∫ (((2t^2 +2))/t^5 )Σ_(k=0) ^(6 ) C_6 ^k t^k (−1)^(6−k) ...be continued...$
	$\Rightarrow (1 - t) x = (1 + t) i \sqrt{2} \Rightarrow x = i \sqrt{2} \times \frac{1 + t}{1 - t}$ $\frac{d x}{d t} = i \sqrt{2} \times \frac{1 - t + (1 + t)}{{(1 - t)}^{2}} = \frac{2 i \sqrt{2}}{{(1 - t)}^{2}}$ $x + i \sqrt{2} = i \sqrt{2 (} \frac{1 + t}{1 - t} + 1)$ $= i \sqrt{2} \times \frac{2}{1 - t} = \frac{2 i \sqrt{2}}{1 - t} \Rightarrow$ $I = \int \frac{- 2 {(\frac{1 + t}{1 - t})}^{2} - 2}{t^{5} {(2 i \sqrt{2})}^{10} {(\frac{1}{1 - t})}^{10}} \times \frac{2 i \sqrt{2}}{{(1 - t)}^{2}}$ $= - 2 \frac{1}{{(2 i \sqrt{2})}^{9}} \int \frac{{(\frac{1 + t}{1 - t})}^{2} + 1}{t^{5} {(1 - t)}^{2}} \times {(1 - t)}^{10} d t$ $= - \frac{2}{{(2 i \sqrt{2})}^{9}} \int \frac{{(1 - t)}^{8} {\frac{{(1 + t)}^{2}}{{(1 - t)}^{2}} + 1}}{t^{5}} d t$ $= - \frac{2}{{(2 i \sqrt{2})}^{9}} \int \frac{{(1 - t)}^{8} ({(1 + t)}^{2} + {(1 - t)}^{2})}{t^{5} {(1 - t)}^{2}} d t$ $= \frac{- 2}{{(2 i \sqrt{2})}^{9}} \int \frac{{(1 - t)}^{6} (t^{2} + 2 t + 1 + t^{2} - 2 t + 1)}{t^{5}} d t$ $= \frac{- 2}{{(2 i \sqrt{2})}^{9}} \int \frac{(2 t^{2} + 2)}{t^{5}} \sum_{k = 0}^{6} C_{6}^{k} t^{k} {(- 1)}^{6 - k}$ $. . . b e c o n t i n u e d . . .$
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