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


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Question Number 120599 by MagdiRagheb last updated on 01/Nov/20

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