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t-7-ln-t-dt-




Question Number 135257 by leena12345 last updated on 11/Mar/21
∫t^7 ln(t)dt
$$\int{t}^{\mathrm{7}} {ln}\left({t}\right){dt} \\ $$
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
$${I}\left({a}\right)=\int{t}^{{a}} {dt}=\frac{{t}^{{a}+\mathrm{1}} }{{a}+\mathrm{1}}+{C} \\ $$$$\frac{\partial}{\partial{a}}{I}\left({a}\right)=\int{t}^{{a}} {log}\left({t}\right){dt}=\frac{{t}^{{a}+\mathrm{1}} {log}\left({a}+\mathrm{1}\right)}{{a}+\mathrm{1}}−\frac{{t}^{{a}+\mathrm{1}} }{\left({a}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\int{t}^{{a}} {log}\left({t}\right){dt}=\frac{{t}^{{a}+\mathrm{1}} {log}\left({t}\right)}{{a}+\mathrm{1}}−\frac{{t}^{{a}+\mathrm{1}} }{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }+{C} \\ $$$${a}=\mathrm{7}\:\:{then}\:{it}\:{will}\:{be}\:\frac{{t}^{\mathrm{8}} }{\mathrm{8}}{log}\left({t}\right)−\frac{{t}^{\mathrm{8}} }{\mathrm{64}}+{C} \\ $$
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
$${u}={lnt}\:\:\:\:\:\:\:{du}=\frac{\mathrm{1}}{{t}}{dt} \\ $$$$\int{dv}=\int{t}^{\mathrm{7}} {dt}\:\Rightarrow{v}=\frac{\mathrm{1}}{\mathrm{8}}{t}^{\mathrm{8}} \\ $$$$\Rightarrow{u}\centerdot{v}−\int{vdu} \\ $$$$\Rightarrow{lnt}\centerdot\frac{\mathrm{1}}{\mathrm{8}}{t}^{\mathrm{8}} −\int\frac{\mathrm{1}}{\mathrm{8}}{t}^{\mathrm{8}} \centerdot\frac{{dt}}{{t}} \\ $$$$\Rightarrow\frac{{t}^{\mathrm{8}} }{\mathrm{8}}\centerdot{lnt}−\frac{\mathrm{1}}{\mathrm{8}}\int{t}^{\mathrm{7}} {dt} \\ $$$$\Rightarrow\frac{{t}^{\mathrm{8}} }{\mathrm{8}}\centerdot{lnt}−\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\mathrm{1}}{\mathrm{8}}{t}^{\mathrm{8}} +{c} \\ $$$$\Rightarrow\frac{{t}^{\mathrm{8}} }{\mathrm{8}}\centerdot{lnt}−\frac{{t}^{\mathrm{8}} }{\mathrm{64}}+{c} \\ $$

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