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Question Number 135257 by leena12345 last updated on 11/Mar/21

∫t^7 ln(t)dt

∫t7ln(t)dt

Commented by leena12345 last updated on 11/Mar/21

help

help

Answered by Dwaipayan Shikari last updated on 11/Mar/21

I(a)=∫t^a dt=(t^(a+1) /(a+1))+C (∂/∂a)I(a)=∫t^a log(t)dt=((t^(a+1) log(a+1))/(a+1))−(t^(a+1) /((a+1)^2 )) ∫t^a log(t)dt=((t^(a+1) log(t))/(a+1))−(t^(a+1) /((a+1)^2 ))+C a=7 then it will be (t^8 /8)log(t)−(t^8 /(64))+C

I(a)=∫tadt=ta+1a+1+C∂∂aI(a)=∫talog(t)dt=ta+1log(a+1)a+1−ta+1(a+1)2∫talog(t)dt=ta+1log(t)a+1−ta+1(a+1)2+Ca=7thenitwillbet88log(t)−t864+C

Answered by 0731619177 last updated on 11/Mar/21

u=lnt du=(1/t)dt ∫dv=∫t^7 dt ⇒v=(1/8)t^8 ⇒u∙v−∫vdu ⇒lnt∙(1/8)t^8 −∫(1/8)t^8 ∙(dt/t) ⇒(t^8 /8)∙lnt−(1/8)∫t^7 dt ⇒(t^8 /8)∙lnt−(1/8)∙(1/8)t^8 +c ⇒(t^8 /8)∙lnt−(t^8 /(64))+c

u=lntdu=1tdt∫dv=∫t7dt⇒v=18t8⇒u⋅v−∫vdu⇒lnt⋅18t8−∫18t8⋅dtt⇒t88⋅lnt−18∫t7dt⇒t88⋅lnt−18⋅18t8+c⇒t88⋅lnt−t864+c

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