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Let x = u^6 dx = 6u^5 du I = ∫(u^3 /((1+u^2 )^2 )) ×6u^5 du =6 ∫(u^8 /(1+2u^2 +u^4 )) du =6 ∫[((−4)/(u^2 +1))+(1/((1+u^2 )^2 ))+u^4 −2u^2 +3]du =6[−4tan^(−1) (u) + I_1 + (u^5 /5) − (2/3)u^3 +3u]+c I_1 = ∫(1/((1+u^2 )^2 )) du Put u = tan z du = sec^2 z dz I_1 = ∫(1/(sec^4 z))×sec^2 z dz = ∫cos^2 z dz = ∫(1/2)(cos 2z + 1)dz = (1/2)((1/2) sin 2z + z) = (1/2)×(u/(1+u^2 )) + (1/2) tan^(−1) u I = −4 tan^(−1) u + (u/(2(1+u^2 ))) + (1/2) tan^(−1) u +(u^5 /5) − (2/3)u^3 + 3u + c = −(7/2) tan^(−1) u + (u/(2(1+u^2 ))) + (1/5)u^5 + 3u + c = −(7/2) tan^(−1) ( ^6 (√x)) + ((x)^(1/6) /(2(1+(x^2 )^(1/6) ))) + (1/5) (x^5 )^(1/6) + 3 (x)^(1/6) + c

$L e t x = u^{6} d x = 6 u^{5} d u$ $I = \int \frac{u^{3}}{{(1 + u^{2})}^{2}} \times 6 u^{5} d u$ $= 6 \int \frac{u^{8}}{1 + 2 u^{2} + u^{4}} d u$ $= 6 \int [\frac{- 4}{u^{2} + 1} + \frac{1}{{(1 + u^{2})}^{2}} + u^{4} - 2 u^{2} + 3] d u$ $= 6 [- 4 {t a n}^{- 1} (u) + I_{1} + \frac{u^{5}}{5} - \frac{2}{3} u^{3} + 3 u] + c$ $I_{1} = \int \frac{1}{{(1 + u^{2})}^{2}} d u$ $P u t u = t a n z d u = {s e c}^{2} z d z$ $I_{1} = \int \frac{1}{{s e c}^{4} z} \times {s e c}^{2} z d z = \int {c o s}^{2} z d z$ $= \int \frac{1}{2} (c o s 2 z + 1) d z$ $= \frac{1}{2} (\frac{1}{2} s i n 2 z + z) = \frac{1}{2} \times \frac{u}{1 + u^{2}} + \frac{1}{2} {t a n}^{- 1} u$ $I = - 4 {t a n}^{- 1} u + \frac{u}{2 (1 + u^{2})} + \frac{1}{2} {t a n}^{- 1} u + \frac{u^{5}}{5} - \frac{2}{3} u^{3} + 3 u + c$ $= - \frac{7}{2} {t a n}^{- 1} u + \frac{u}{2 (1 + u^{2})} + \frac{1}{5} u^{5} + 3 u + c$ $= - \frac{7}{2} {t a n}^{- 1} (\overset{6}{} \sqrt{x}) + \frac{\sqrt[6]{x}}{2 (1 + \sqrt[6]{x^{2}})} + \frac{1}{5} \sqrt[6]{x^{5}} + 3 \sqrt[6]{x} + c$

Question Number 120582 Answers: 0 Comments: 0

Let u=x^(3/5) du = (3/(5x^(2/5) )) I = (5/3)∫(u/( (√(3−2u)))) du = −(5/6) ∫((3−2u−3)/( (√(3−2u)))) du = −(5/3) ∫[(√(3−2u)) −3(3−2u)^(−(1/2)) ] du = −(5/3)[−(1/3)(3−2u)^(3/2) +3(3−2u)^(1/2) ]+c = −(5/(18))(3−2u)^(1/2) [−3+2u + 9]+c = −(5/(18))(3−2u)^(1/2) (6+2u)+c = −(5/9)(√(3−2x^(3/5) ))(3+x^(3/5) )+c

$L e t u = x^{\frac{3}{5}} d u = \frac{3}{5 x^{\frac{2}{5}}}$ $I = \frac{5}{3} \int \frac{u}{\sqrt{3 - 2 u}} d u = - \frac{5}{6} \int \frac{3 - 2 u - 3}{\sqrt{3 - 2 u}} d u$ $= - \frac{5}{3} \int [\sqrt{3 - 2 u} - 3 {(3 - 2 u)}^{- \frac{1}{2}}] d u$ $= - \frac{5}{3} [- \frac{1}{3} {(3 - 2 u)}^{\frac{3}{2}} + 3 {(3 - 2 u)}^{\frac{1}{2}}] + c$ $= - \frac{5}{18} {(3 - 2 u)}^{\frac{1}{2}} [- 3 + 2 u + 9] + c$ $= - \frac{5}{18} {(3 - 2 u)}^{\frac{1}{2}} (6 + 2 u) + c$ $= - \frac{5}{9} \sqrt{3 - 2 x^{\frac{3}{5}}} (3 + x^{\frac{3}{5}}) + c$

Question Number 120562 Answers: 1 Comments: 0

Question Number 120554 Answers: 0 Comments: 0

Prove that for all a>0 ∫_([−a;a]) arg(Γ((1/2) −ix))dx =0 Deduce that f: x→arg(Γ((1/2) −ix)) is an old function on R

$P r o v e t h a t f o r a l l a > 0$ $\int_{[- a; a]} a r g (Γ (\frac{1}{2} - i x)) d x = 0$ $D e d u c e t h a t$ $f : x \to a r g (Γ (\frac{1}{2} - i x)) i s a n o l d f u n c t i o n o n R$

Question Number 120552 Answers: 0 Comments: 0

evaluate: ∫_0 ^( ∞) ((x/(x+1)))^(x!) dx

$evaluate : \int_{0}^{\infty} {(\frac{x}{x + 1})}^{x!} d x$

Question Number 120549 Answers: 3 Comments: 0

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