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Question Number 126994 by bramlexs22 last updated on 26/Dec/20
∫x41+x8dx=?
Answered by liberty last updated on 26/Dec/20
Ω=∫x41+x8dx=∫dxx4+1x4=∫dx(x2+1x2)2−2Ω=∫dx(x2+1x2+2)(x2+1x2−2)Ω=24∫(dx(x2+1x2−2)−dx(x2+1x2+2))Ω=28[∫(1−1x2)+(1+1x2)x2+1x2−2dx−∫(1−1x2)+(1+1x2)x2+1x2+2dx]Ω=28[∫d(x+1x)(x+1x)2−(2+2)+∫d(x−1x)(x−1x)2+(2−2)]−28[∫d(x+1x)(x+1x)2−(2−2)+∫d(x−1x)(x−1x)2+(2+2)]Ω=28[122+2ln∣x2−x2+2+1x2+x2+2+1∣+12−2tan−1(x2−1x2−2)−122−2ln∣x2−x2−2+1x2+x2−2+1∣−12−2tan−1(x2+1x2+2)]+c
Commented by bramlexs22 last updated on 26/Dec/20
waw....
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