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Question Number 156422 by SANOGO last updated on 11/Oct/21

Commented by cortano last updated on 11/Oct/21

 let x=t^6  →dx=6t^5  dt  I=∫ (t^3 /(1−t^2 )) (6t^5  dt)  = ∫ ((6t^8 )/(1−t^2 )) dt   partial fraction

letx=t6dx=6t5dtI=t31t2(6t5dt)=6t81t2dtpartialfraction

Commented by SANOGO last updated on 11/Oct/21

merci bien

mercibien

Answered by puissant last updated on 11/Oct/21

K=∫(1/( (√x)−(x)^(1/3) ))dx ; x=u^6 →dx=6u^5 du  ⇒ K=∫((6u^5 )/(u^3 −u^2 ))du = 6∫(u^3 /(u−1))du  =6∫((u^3 −1+1)/(u−1))du = 6∫((u^3 −1)/(u−1))du+6∫(1/(u−1))du  =6∫{u^2 +u+1}du+6ln∣u∣  ⇒ K=2u^3 +3u^2 +6u+6ln∣u∣+C  ⇒ K=2(x^(1/6) )^3 +3(x^(1/6) )^2 +6x^(1/6) +6ln∣x^(1/6) ∣+C    ∴∵ K=2(√x)+3(x)^(1/3) +6(x)^(1/6) +ln∣x∣+C..                   ..........Le puissant...........

K=1xx3dx;x=u6dx=6u5duK=6u5u3u2du=6u3u1du=6u31+1u1du=6u31u1du+61u1du=6{u2+u+1}du+6lnuK=2u3+3u2+6u+6lnu+CK=2(x16)3+3(x16)2+6x16+6lnx16+C∴∵K=2x+3x3+6x6+lnx+C............Lepuissant...........

Commented by SANOGO last updated on 11/Oct/21

merci bien

mercibien

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