(1a) ∫(dx/(e^(5x) +e^(3x) +e^(2x) ))=? (1b) ∫(dx/(e^(5x) −e^(3x) −e^(2x) ))=? (2a) ∫(dx/(x^(5/3) +x^(3/5) ))=? (2b) ∫(dx/(x^(5/3) −x^(3/5) ))=?


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Question Number 46323 by MJS last updated on 24/Oct/18

$(1 a) \int \frac{d x}{e^{5 x} + e^{3 x} + e^{2 x}} = ?$ $(1 b) \int \frac{d x}{e^{5 x} - e^{3 x} - e^{2 x}} = ?$ $(2 a) \int \frac{d x}{x^{\frac{5}{3}} + x^{\frac{3}{5}}} = ?$ $(2 b) \int \frac{d x}{x^{\frac{5}{3}} - x^{\frac{3}{5}}} = ?$
Commented by maxmathsup by imad last updated on 24/Oct/18
$(2a) we use the changement x=t^(15) ⇒ I = ∫ ((15 t^(14) dt)/((t^(15) )^(5/3) +(t^(15) )^(3/5) )) = ∫ ((15 t^(14) )/(t^(25) + t^9 )) dt =15 ∫ (t^5 /(t^(16) +1)) dt but t^(16) +1 =(t^8 )^2 +1 =(t^8 +1)^2 −2t^8 =(t^8 +1−(√2)t^4 )(t^8 +1+(√2)t^4 ) ⇒ (1/((t^8 −(√2)t^4 +1)(t^8 +(√2)t^4 +1))) =(1/λ){ (1/(t^8 +(√2)t^4 +1)) −(1/(t^8 −(√2)t^4 +1))} λ=t^8 −(√2)t^4 +1−t^8 −(√2)t^4 −1 = −2(√2)t^4 ⇒ F(x)=(t^5 /(t^(16) +1)) =−(1/(2(√2))){ (t/(t^8 +(√2)t^4 +1)) −(t/(t^8 −(√2)t^4 +1))} ⇒ I =−((15)/(2(√2))) { ∫ ((tdt)/(t^8 +(√2)t^4 +1)) −∫ ((tdt)/(t^8 −(√2) t^4 +1))} also t^8 +(√2)t^4 +1 =u^2 +(√2)u +1 (u=t^4 ) =u^2 +2 ((√2)/2)u +(1/2) +(1/2) =(u +((√2)/2))^2 +(1/2) =(t^4 +((√2)/2))^2 +(1/2) ⇒ ∫ ((t dt)/(t^8 +(√2)t^4 +1)) =∫ (t/((t^4 +((√2)/2))^2 +(1/2))) dt changement t^4 =α give t =α^(1/4) ∫ (....)dt = ∫ (α^(1/4) /((α+((√2)/2))^2 +(1/2))) (1/4) α^((1/4)−1) dα = ∫ (dα/((√α){ (α+((√2)/2))^2 +(1/2)})) ....be continued....$
$(2 a) w e u s e t h e c h a n g e m e n t x = t^{15} \Rightarrow$ $I = \int \frac{15 t^{14} d t}{{(t^{15})}^{\frac{5}{3}} + {(t^{15})}^{\frac{3}{5}}} = \int \frac{15 t^{14}}{t^{25} + t^{9}} d t = 15 \int \frac{t^{5}}{t^{16} + 1} d t b u t$ $t^{16} + 1 = {(t^{8})}^{2} + 1 = {(t^{8} + 1)}^{2} - 2 t^{8} = (t^{8} + 1 - \sqrt{2} t^{4}) (t^{8} + 1 + \sqrt{2} t^{4}) \Rightarrow$ $\frac{1}{(t^{8} - \sqrt{2} t^{4} + 1) (t^{8} + \sqrt{2} t^{4} + 1)} = \frac{1}{λ} {\frac{1}{t^{8} + \sqrt{2} t^{4} + 1} - \frac{1}{t^{8} - \sqrt{2} t^{4} + 1}}$ $λ = t^{8} - \sqrt{2} t^{4} + 1 - t^{8} - \sqrt{2} t^{4} - 1 = - 2 \sqrt{2} t^{4} \Rightarrow$ $F (x) = \frac{t^{5}}{t^{16} + 1} = - \frac{1}{2 \sqrt{2}} {\frac{t}{t^{8} + \sqrt{2} t^{4} + 1} - \frac{t}{t^{8} - \sqrt{2} t^{4} + 1}} \Rightarrow$ $I = - \frac{15}{2 \sqrt{2}} {\int \frac{t d t}{t^{8} + \sqrt{2} t^{4} + 1} - \int \frac{t d t}{t^{8} - \sqrt{2} t^{4} + 1}} a l s o$ $t^{8} + \sqrt{2} t^{4} + 1 = u^{2} + \sqrt{2} u + 1 (u = t^{4})$ $= u^{2} + 2 \frac{\sqrt{2}}{2} u + \frac{1}{2} + \frac{1}{2} = {(u + \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2} = {(t^{4} + \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2} \Rightarrow$ $\int \frac{t d t}{t^{8} + \sqrt{2} t^{4} + 1} = \int \frac{t}{{(t^{4} + \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2}} d t c h a n g e m e n t t^{4} = α g i v e t = α^{\frac{1}{4}}$ $\int (. . . .) d t = \int \frac{α^{\frac{1}{4}}}{{(α + \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2}} \frac{1}{4} α^{\frac{1}{4} - 1} d α$ $= \int \frac{d α}{\sqrt{α} {{(α + \frac{\sqrt{2}}{2})}^{2} + \frac{1}{2}}} . . . . b e c o n t i n u e d . . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 24/Oct/18
	$2a)∫(dx/(x^a +x^b )) [here ab=1 b=(5/3) a=(3/5)] ∫(dx/(x^b (x^(a−b) +1))) ∫x^(−b) (1−x^t +x^(2t) −x^(3t) +...)dx [ t=a−b = ∫{x^(−b) −x^(t−b) +x^(2t−b) −x^(3t−b) +.. }dx =(x^(−b+1) /(−b+1))−(x^(t−b+1) /(t−b+1))+(x^(2t−b+1) /(2t−b+1))−.... pls check may i proceed this way...$
	$2 a) \int \frac{d x}{x^{a} + x^{b}} [h e r e a b = 1 b = \frac{5}{3} a = \frac{3}{5}]$ $\int \frac{d x}{x^{b} (x^{a - b} + 1)}$ $\int x^{- b} (1 - x^{t} + x^{2 t} - x^{3 t} + . . .) d x [t = a - b =$ $\int {x^{- b} - x^{t - b} + x^{2 t - b} - x^{3 t - b} + . .} d x$ $= \frac{x^{- b + 1}}{- b + 1} - \frac{x^{t - b + 1}}{t - b + 1} + \frac{x^{2 t - b + 1}}{2 t - b + 1} - . . . .$ $p l s c h e c k m a y i p r o c e e d t h i s w a y . . .$
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