Menu Close

calculate-0-dx-2x-1-4-x-3-5-

Tinku Tara 0 Comments
Pin




Question Number 146196 by mathmax by abdo last updated on 11/Jul/21
calculate ∫_0 ^∞ (dx/((2x+1)^4 (x+3)^5 ))
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{dx}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{4}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{5}} } \\ $$
Answered by mathmax by abdo last updated on 12/Jul/21
Ψ=∫_0 ^∞ (dx/((2x+1)^4 (x+3)^5 )) ⇒Ψ=∫_0 ^∞ (dx/((((2x+1)/(x+3)))^4 (x+3)^9 )) we do the changement ((2x+1)/(x+3))=t ⇒2x+1=tx+3t ⇒(2−t)x=3t−1 ⇒x=((3t−1)/(2−t)) ⇒(dx/dt)=((3(2−t)−(3t−1)(−1))/((2−t)^2 ))=((6−3t+3t−1)/((2−t)^2 ))=(5/((2−t)^2 )) x+3=((3t−1)/(2−t))+3=((3t−1+6−3t)/(2−t))=(5/(2−t)) ⇒ Ψ=∫_(1/3) ^2 (5/((2−t)^2 t^4 ((5/(2−t)))^9 ))dt =(1/5^8 )∫_(1/3) ^2 (((2−t)^7 )/t^4 )dt =(1/5^8 )∫_(1/3) ^2 ((Σ_(k=0) ^7 C_7 ^k (−t)^k 2^(7−k) )/t^4 )dt =(2^7 /5^8 )∫_(1/3) ^2 Σ_(k=0) ^7 C_7 ^k (−1)^k t^(k−4) 2^(−k) dt =(2^7 /5^8 )Σ_(k=0) ^7 (−1)^k 2^(−k) C_7 ^k ∫_(1/3) ^2 t^(k−4) dt =(2^7 /5^8 ){Σ_(k=0 and k≠3) ^7 (−(1/2))^k C_7 ^k [(1/(k−3))t^(k−3) ]_(1/3) ^2 −(1/8)C_7 ^3 [log∣t∣]_(1/3) ^2 } Ψ=(2^7 /5^8 )Σ_(k=0and k≠3) ^7 (−(1/2))^k (C_7 ^k /(k−3))(2^(k−3) −(1/3^(k−3) )) −(2^7 /(8.5^8 ))C_7 ^3 (log2+log3)
$$\Psi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{4}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{5}} }\:\Rightarrow\Psi=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\left(\frac{\mathrm{2x}+\mathrm{1}}{\mathrm{x}+\mathrm{3}}\right)^{\mathrm{4}} \left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{9}} } \\ $$$$\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\frac{\mathrm{2x}+\mathrm{1}}{\mathrm{x}+\mathrm{3}}=\mathrm{t}\:\Rightarrow\mathrm{2x}+\mathrm{1}=\mathrm{tx}+\mathrm{3t}\:\Rightarrow\left(\mathrm{2}−\mathrm{t}\right)\mathrm{x}=\mathrm{3t}−\mathrm{1} \\ $$$$\Rightarrow\mathrm{x}=\frac{\mathrm{3t}−\mathrm{1}}{\mathrm{2}−\mathrm{t}}\:\Rightarrow\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{3}\left(\mathrm{2}−\mathrm{t}\right)−\left(\mathrm{3t}−\mathrm{1}\right)\left(−\mathrm{1}\right)}{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} }=\frac{\mathrm{6}−\mathrm{3t}+\mathrm{3t}−\mathrm{1}}{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} }=\frac{\mathrm{5}}{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} } \\ $$$$\mathrm{x}+\mathrm{3}=\frac{\mathrm{3t}−\mathrm{1}}{\mathrm{2}−\mathrm{t}}+\mathrm{3}=\frac{\mathrm{3t}−\mathrm{1}+\mathrm{6}−\mathrm{3t}}{\mathrm{2}−\mathrm{t}}=\frac{\mathrm{5}}{\mathrm{2}−\mathrm{t}}\:\Rightarrow \\ $$$$\Psi=\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\frac{\mathrm{5}}{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{2}} \mathrm{t}^{\mathrm{4}} \left(\frac{\mathrm{5}}{\mathrm{2}−\mathrm{t}}\right)^{\mathrm{9}} }\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\frac{\left(\mathrm{2}−\mathrm{t}\right)^{\mathrm{7}} }{\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{8}} }\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\frac{\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \:\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} \left(−\mathrm{t}\right)^{\mathrm{k}} \:\mathrm{2}^{\mathrm{7}−\mathrm{k}} }{\mathrm{t}^{\mathrm{4}} }\mathrm{dt} \\ $$$$=\frac{\mathrm{2}^{\mathrm{7}} }{\mathrm{5}^{\mathrm{8}} }\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \:\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} \:\left(−\mathrm{1}\right)^{\mathrm{k}} \:\mathrm{t}^{\mathrm{k}−\mathrm{4}} \mathrm{2}^{−\mathrm{k}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{2}^{\mathrm{7}} }{\mathrm{5}^{\mathrm{8}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \:\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{2}^{−\mathrm{k}} \:\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} \:\int_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:\mathrm{t}^{\mathrm{k}−\mathrm{4}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{2}^{\mathrm{7}} }{\mathrm{5}^{\mathrm{8}} }\left\{\sum_{\mathrm{k}=\mathrm{0}\:\mathrm{and}\:\mathrm{k}\neq\mathrm{3}} ^{\mathrm{7}} \:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{k}} \:\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} \:\left[\frac{\mathrm{1}}{\mathrm{k}−\mathrm{3}}\mathrm{t}^{\mathrm{k}−\mathrm{3}} \right]_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \:−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{C}_{\mathrm{7}} ^{\mathrm{3}} \:\left[\mathrm{log}\mid\mathrm{t}\mid\right]_{\frac{\mathrm{1}}{\mathrm{3}}} ^{\mathrm{2}} \right\} \\ $$$$\Psi=\frac{\mathrm{2}^{\mathrm{7}} }{\mathrm{5}^{\mathrm{8}} }\sum_{\mathrm{k}=\mathrm{0and}\:\mathrm{k}\neq\mathrm{3}} ^{\mathrm{7}} \:\:\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{k}} \:\frac{\mathrm{C}_{\mathrm{7}} ^{\mathrm{k}} }{\mathrm{k}−\mathrm{3}}\left(\mathrm{2}^{\mathrm{k}−\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{k}−\mathrm{3}} }\right) \\ $$$$−\frac{\mathrm{2}^{\mathrm{7}} }{\mathrm{8}.\mathrm{5}^{\mathrm{8}} }\mathrm{C}_{\mathrm{7}} ^{\mathrm{3}} \:\left(\mathrm{log2}+\mathrm{log3}\right) \\ $$
Answered by Olaf_Thorendsen last updated on 12/Jul/21
R(x) = (1/((2x+1)^4 (x+3)^5 )) R(x) = (((32)/(3125))/((2x+1)^4 ))−(((32)/(3125))/((2x+1)^3 ))+(((96)/(15625))/((2x+1)^2 )) −(((224)/(78125))/(2x+1))+((1/(625))/((x+3)^5 ))+((8/(3125))/((x+3)^4 ))+((8/(3125))/((x+3)^3 )) +(((32)/(15625))/((x+2)^2 ))+(((112)/(78125))/(x+3)) ∫R(x)dx = −(((16)/(9375))/((2x+1)^3 ))+((8/(3125))/((2x+1)^2 )) −(((48)/(15625))/(2x+1))−((112)/(78125))ln(2x+1)−((1/(2500))/((x+3)^4 )) −((8/(9375))/((x+3)^3 ))−((4/(3125))/((x+3)^2 ))−(((32)/(15625))/(x+3)) +((112)/(78125))ln(x+3) Ω = ∫_0 ^∞ R(x)dx = ((15593)/(5062500))−((112)/(78125))ln6
$$\mathrm{R}\left({x}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{4}} \left({x}+\mathrm{3}\right)^{\mathrm{5}} } \\ $$$$\mathrm{R}\left({x}\right)\:=\:\frac{\frac{\mathrm{32}}{\mathrm{3125}}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{4}} }−\frac{\frac{\mathrm{32}}{\mathrm{3125}}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} }+\frac{\frac{\mathrm{96}}{\mathrm{15625}}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$−\frac{\frac{\mathrm{224}}{\mathrm{78125}}}{\mathrm{2}{x}+\mathrm{1}}+\frac{\frac{\mathrm{1}}{\mathrm{625}}}{\left({x}+\mathrm{3}\right)^{\mathrm{5}} }+\frac{\frac{\mathrm{8}}{\mathrm{3125}}}{\left({x}+\mathrm{3}\right)^{\mathrm{4}} }+\frac{\frac{\mathrm{8}}{\mathrm{3125}}}{\left({x}+\mathrm{3}\right)^{\mathrm{3}} } \\ $$$$+\frac{\frac{\mathrm{32}}{\mathrm{15625}}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }+\frac{\frac{\mathrm{112}}{\mathrm{78125}}}{{x}+\mathrm{3}} \\ $$$$\int\mathrm{R}\left({x}\right){dx}\:=\:−\frac{\frac{\mathrm{16}}{\mathrm{9375}}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{3}} }+\frac{\frac{\mathrm{8}}{\mathrm{3125}}}{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$−\frac{\frac{\mathrm{48}}{\mathrm{15625}}}{\mathrm{2}{x}+\mathrm{1}}−\frac{\mathrm{112}}{\mathrm{78125}}\mathrm{ln}\left(\mathrm{2}{x}+\mathrm{1}\right)−\frac{\frac{\mathrm{1}}{\mathrm{2500}}}{\left({x}+\mathrm{3}\right)^{\mathrm{4}} } \\ $$$$−\frac{\frac{\mathrm{8}}{\mathrm{9375}}}{\left({x}+\mathrm{3}\right)^{\mathrm{3}} }−\frac{\frac{\mathrm{4}}{\mathrm{3125}}}{\left({x}+\mathrm{3}\right)^{\mathrm{2}} }−\frac{\frac{\mathrm{32}}{\mathrm{15625}}}{{x}+\mathrm{3}} \\ $$$$+\frac{\mathrm{112}}{\mathrm{78125}}\mathrm{ln}\left({x}+\mathrm{3}\right) \\ $$$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\infty} \mathrm{R}\left({x}\right){dx}\:=\:\frac{\mathrm{15593}}{\mathrm{5062500}}−\frac{\mathrm{112}}{\mathrm{78125}}\mathrm{ln6} \\ $$
Commented by mathmax by abdo last updated on 12/Jul/21
thank you sir.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *