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Question Number 41989 by Raj Singh last updated on 16/Aug/18

Commented by maxmathsup by imad last updated on 16/Aug/18

$l e t I = \int \frac{d x}{{(x^{2} + 2)}^{3}} c h a n g e m e n t x = \sqrt{2} t a n (t) g i v e$ $I = \int \frac{\sqrt{2} (1 + {t a n}^{2} t)}{8 {(1 + {t a n}^{2} t)}^{3}} d t = \frac{\sqrt{2}}{8} \int {c o s}^{4} t d t$ $= \frac{\sqrt{2}}{8} \int {(\frac{1 + c o s (2 t)}{2})}^{2} d t = \frac{\sqrt{2}}{32} \int (1 + 2 c o s (2 t) + {c o s}^{2} (2 t)) d t$ $= \frac{\sqrt{2}}{32} t + \frac{\sqrt{2}}{16} \int c o s (2 t) d t + \frac{\sqrt{2}}{64} \int (1 + c o s (4 t)) d t$ $= \frac{\sqrt{2}}{32} t + \frac{\sqrt{2}}{32} s i n (2 t) + \frac{\sqrt{2}}{64} t + \frac{\sqrt{2}}{4 \times 64} s i n (4 t)$ $= \frac{3 \sqrt{2}}{32} t + \frac{\sqrt{2}}{32} s i n (2 t) + \frac{\sqrt{2}}{256} s i n (4 t) b u t t = a r c t a n (\frac{x}{\sqrt{2}})$ $s i n (2 t) = \frac{2 t a n (t)}{1 + {t a n}^{2} t} = \frac{2 \frac{x}{\sqrt{2}}}{1 + {(\frac{x}{\sqrt{2}})}^{2}} = \frac{2 x \sqrt{2}}{2 + x^{2}}$ $s i n (4 t) = 2 s i n (2 t) c o s (2 t) = \frac{4 t a n (t)}{1 + {t a n}^{2} (t)} \frac{1 - {t a n}^{2} (t)}{1 + {t a n}^{2} t}$ $= \frac{4 \frac{x}{\sqrt{2}} (1 - {(\frac{x}{\sqrt{2}})}^{2})}{{(1 + {(\frac{x}{\sqrt{2}})}^{2})}^{2}} = \frac{\frac{4 x}{\sqrt{2}} (\frac{2 - x^{2}}{2})}{{(\frac{2 + x^{2}}{2})}^{2}} = 4 . \frac{4}{2 \sqrt{2}} \frac{2 - x^{2}}{{(2 + x^{2})}^{2}} = \frac{8}{\sqrt{2}} \frac{2 - x^{2}}{{(2 + x^{2})}^{2}} \Rightarrow$ $I = \frac{3 \sqrt{2}}{32} a r c t a n (\frac{x}{\sqrt{2}}) + \frac{\sqrt{2}}{32} \frac{2 x \sqrt{2}}{2 + x^{2}} + \frac{\sqrt{2}}{256} \frac{8}{\sqrt{2}} \frac{2 - x^{2}}{{(2 + x^{2})}^{2}} + c .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
	$x=(√2) tanα dx=(√2) sec^2 α dα ∫(((√2) sec^2 αdα)/({2(tan^2 α +1)}^3 )) (((√2) )/2^3 )∫((sec^2 α)/(sec^6 α))dα (1/(4(√2)))∫cos^4 αdα (1/(4(√2)))∫((1+cos2α)/2)×((1+cos2α)/2)dα (1/(16(√2) ))∫1+2cos2α+((1+cos4α)/2) dα (1/(16(√2) ))∫dα+(1/(8(√2) ))∫cos2α+(1/(32(√2)))∫dα+(1/(32(√2)))∫cos4αdα =(1/(16(√2)))α+(1/(8(√2)))×((sin2α)/2)+(1/(32(√2)))α+(1/(32(√2)))×((sin4α)/4)+c =(1/(32(√2) ))(2α+α)+(1/(16(√(2 ))))×((2tanα)/(1+tan^2 α))+(1/(128(√2)))×(2.((2tanα)/(1+tan^2 α)).((1−tan^2 α)/(1+tan^2 α))) =(1/(32(√2)))×3tan^(−1) ((x/(√2)))+(1/(8(√2)))×((x/(√2))/(1+(x^2 /2)))+(1/(32(√2)))(((x/(√2))/(1+(x^2 /2))).((1−(x^2 /2))/(1+(x^2 /2)))) =(1/(32(√2)))×3tan^(−1) ((x/(√2)))+(1/(8(√2)))×(2/(√2))×(x/(2+x^2 ))+(1/(32(√2)))((√2) .(x/(2+x^2 )).((2−x^2 )/(2+x^2 ))) =(1/(32(√2)))×3tan^(−1) ((x/((√2) )))+(1/8).(x/(2+x^2 ))+(1/(32)){((x.(2−x^2 ))/((2+x^2 )^2 ))} =(3/(32(√2)))tan^(−1) ((x/(√2)))+(1/8).(x/(2+x^2 )).(1+(1/4).((2−x^2 )/(2+x^2 ))) =((3(√2))/(64))tan^(−1) (((x(√2))/2))+(1/8).(x/(2+x^2 ))(((8+4x^2 +2−x^2 )/(8+4x^2 ))) =((3(√2))/(64))tan^(−1) (((x(√2))/2))+(1/8).(x/(2+x^2 )).(((10+3x^2 )/(8+4x^2 ))) ((3(√2))/(64))tan^(−1) (((x(√2))/2))+((x(10+3x^2 ))/((16+8x^2 )(8+4x^2 )))$
	$x = \sqrt{2} t a n α d x = \sqrt{2} {s e c}^{2} α d α$ $\int \frac{\sqrt{2} {s e c}^{2} α d α}{{2 ({t a n}^{2} α + 1)}^{3}}$ $\frac{\sqrt{2}}{2^{3}} \int \frac{{s e c}^{2} α}{{s e c}^{6} α} d α$ $\frac{1}{4 \sqrt{2}} \int {c o s}^{4} α d α$ $\frac{1}{4 \sqrt{2}} \int \frac{1 + c o s 2 α}{2} \times \frac{1 + c o s 2 α}{2} d α$ $\frac{1}{16 \sqrt{2}} \int 1 + 2 c o s 2 α + \frac{1 + c o s 4 α}{2} d α$ $\frac{1}{16 \sqrt{2}} \int d α + \frac{1}{8 \sqrt{2}} \int c o s 2 α + \frac{1}{32 \sqrt{2}} \int d α + \frac{1}{32 \sqrt{2}} \int c o s 4 α d α$ $= \frac{1}{16 \sqrt{2}} α + \frac{1}{8 \sqrt{2}} \times \frac{s i n 2 α}{2} + \frac{1}{32 \sqrt{2}} α + \frac{1}{32 \sqrt{2}} \times \frac{s i n 4 α}{4} + c$ $= \frac{1}{32 \sqrt{2}} (2 α + α) + \frac{1}{16 \sqrt{2}} \times \frac{2 t a n α}{1 + {t a n}^{2} α} + \frac{1}{128 \sqrt{2}} \times (2 . \frac{2 t a n α}{1 + {t a n}^{2} α} . \frac{1 - {t a n}^{2} α}{1 + {t a n}^{2} α})$ $= \frac{1}{32 \sqrt{2}} \times 3 {t a n}^{- 1} (\frac{x}{\sqrt{2}}) + \frac{1}{8 \sqrt{2}} \times \frac{\frac{x}{\sqrt{2}}}{1 + \frac{x^{2}}{2}} + \frac{1}{32 \sqrt{2}} (\frac{\frac{x}{\sqrt{2}}}{1 + \frac{x^{2}}{2}} . \frac{1 - \frac{x^{2}}{2}}{1 + \frac{x^{2}}{2}})$ $= \frac{1}{32 \sqrt{2}} \times 3 {t a n}^{- 1} (\frac{x}{\sqrt{2}}) + \frac{1}{8 \sqrt{2}} \times \frac{2}{\sqrt{2}} \times \frac{x}{2 + x^{2}} + \frac{1}{32 \sqrt{2}} (\sqrt{2} . \frac{x}{2 + x^{2}} . \frac{2 - x^{2}}{2 + x^{2}})$ $= \frac{1}{32 \sqrt{2}} \times 3 {t a n}^{- 1} (\frac{x}{\sqrt{2}}) + \frac{1}{8} . \frac{x}{2 + x^{2}} + \frac{1}{32} {\frac{x . (2 - x^{2})}{{(2 + x^{2})}^{2}}}$ $= \frac{3}{32 \sqrt{2}} {t a n}^{- 1} (\frac{x}{\sqrt{2}}) + \frac{1}{8} . \frac{x}{2 + x^{2}} . (1 + \frac{1}{4} . \frac{2 - x^{2}}{2 + x^{2}})$ $= \frac{3 \sqrt{2}}{64} {t a n}^{- 1} (\frac{x \sqrt{2}}{2}) + \frac{1}{8} . \frac{x}{2 + x^{2}} (\frac{8 + 4 x^{2} + 2 - x^{2}}{8 + 4 x^{2}})$ $= \frac{3 \sqrt{2}}{64} {t a n}^{- 1} (\frac{x \sqrt{2}}{2}) + \frac{1}{8} . \frac{x}{2 + x^{2}} . (\frac{10 + 3 x^{2}}{8 + 4 x^{2}})$ $\frac{3 \sqrt{2}}{64} {t a n}^{- 1} (\frac{x \sqrt{2}}{2}) + \frac{x (10 + 3 x^{2})}{(16 + 8 x^{2}) (8 + 4 x^{2})}$
	Answered by MJS last updated on 16/Aug/18

	$reduction formula :$ $\int \frac{d x}{{({a x}^{2} + b)}^{n}} = \frac{x}{2 b (n - 1) {({a x}^{2} + b)}^{n - 1}} + \frac{2 n - 3}{2 b (n - 1)} \int \frac{d x}{{({a x}^{2} + b)}^{n - 1}}$ $in this case$ $(1) a = 1 b = 2 n = 3$ $\frac{x}{8 {(x^{2} + 2)}^{2}} + \frac{3}{8} \int \frac{d x}{{(x^{2} + 2)}^{2}}$ $(2) a = 1 b = 2 n = 2$ $\frac{x}{8 {(x^{2} + 2)}^{2}} + \frac{3 x}{32 (x^{2} + 2)} + \frac{3}{32} \int \frac{d x}{x^{2} + 2}$ $\int \frac{d x}{x^{2} + c} = \frac{\sqrt{c}}{c} \arctan \frac{x \sqrt{c}}{c}$ $\frac{x}{8 {(x^{2} + 2)}^{2}} + \frac{3 x}{32 (x^{2} + 2)} + \frac{3 \sqrt{2}}{64} \arctan \frac{x \sqrt{2}}{2} + C$ $\frac{x (3 x^{2} + 10)}{32 {(x^{2} + 2)}^{2}} + \frac{3 \sqrt{2}}{64} \arctan \frac{x \sqrt{2}}{2} + C$
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