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Question-121886




Question Number 121886 by oustmuchiya@gmail.com last updated on 12/Nov/20
Commented by TANMAY PANACEA last updated on 12/Nov/20
plz purchase a good text book and study the  basic
$${plz}\:{purchase}\:{a}\:{good}\:{text}\:{book}\:{and}\:{study}\:{the} \\ $$$${basic} \\ $$
Answered by mathmax by abdo last updated on 12/Nov/20
1) ∫_1 ^3 (3x^3  +3)^4  dx =3^4  ∫_1 ^3 (x^3 +1)^4 dx  =3^4  ∫_1 ^3  Σ_(k=0) ^4  C_4 ^k  (x^3 )^k dx =3^k  Σ_(k=0) ^4 C_4 ^k  ∫_1 ^3  x^(3k)  dx  =3^k  Σ_(k=0) ^4 C_4 ^k [(x^(3k+1) /(3k+1))]_1 ^3  =3^k  Σ_(k=0) ^4 (C_4 ^k /(3k+1))(3^(3k+1) −1)
$$\left.\mathrm{1}\right)\:\int_{\mathrm{1}} ^{\mathrm{3}} \left(\mathrm{3x}^{\mathrm{3}} \:+\mathrm{3}\right)^{\mathrm{4}} \:\mathrm{dx}\:=\mathrm{3}^{\mathrm{4}} \:\int_{\mathrm{1}} ^{\mathrm{3}} \left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{4}} \mathrm{dx} \\ $$$$=\mathrm{3}^{\mathrm{4}} \:\int_{\mathrm{1}} ^{\mathrm{3}} \:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{4}} \:\mathrm{C}_{\mathrm{4}} ^{\mathrm{k}} \:\left(\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{k}} \mathrm{dx}\:=\mathrm{3}^{\mathrm{k}} \:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{4}} \mathrm{C}_{\mathrm{4}} ^{\mathrm{k}} \:\int_{\mathrm{1}} ^{\mathrm{3}} \:\mathrm{x}^{\mathrm{3k}} \:\mathrm{dx} \\ $$$$=\mathrm{3}^{\mathrm{k}} \:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{4}} \mathrm{C}_{\mathrm{4}} ^{\mathrm{k}} \left[\frac{\mathrm{x}^{\mathrm{3k}+\mathrm{1}} }{\mathrm{3k}+\mathrm{1}}\right]_{\mathrm{1}} ^{\mathrm{3}} \:=\mathrm{3}^{\mathrm{k}} \:\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{4}} \frac{\mathrm{C}_{\mathrm{4}} ^{\mathrm{k}} }{\mathrm{3k}+\mathrm{1}}\left(\mathrm{3}^{\mathrm{3k}+\mathrm{1}} −\mathrm{1}\right) \\ $$
Answered by mathmax by abdo last updated on 12/Nov/20
let decompose F(x)=((2x+1)/(2x^2 −3x−2))  2x^2 −3x−2=0 →Δ=9−4(−4) =25 ⇒x_1 =((3+5)/4)=2  x_2 =((3−5)/4) =−(1/2) ⇒F(x) =((2x+1)/(2(x−2)(x+(1/2))))=((2x+1)/((2x+1)(x−2)))=(1/(x−2))  ⇒∫_3 ^4  ((2x+1)/(2x^2 −3x−2))dx =∫_3 ^4  (dx/(x−2))=[ln∣x−2∣]_3 ^4  =ln(2)
$$\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{2x}+\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{2}} \\ $$$$\mathrm{2x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{2}=\mathrm{0}\:\rightarrow\Delta=\mathrm{9}−\mathrm{4}\left(−\mathrm{4}\right)\:=\mathrm{25}\:\Rightarrow\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{5}}{\mathrm{4}}=\mathrm{2} \\ $$$$\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{3}−\mathrm{5}}{\mathrm{4}}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{2x}+\mathrm{1}}{\mathrm{2}\left(\mathrm{x}−\mathrm{2}\right)\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\mathrm{2x}+\mathrm{1}}{\left(\mathrm{2x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{x}−\mathrm{2}} \\ $$$$\Rightarrow\int_{\mathrm{3}} ^{\mathrm{4}} \:\frac{\mathrm{2x}+\mathrm{1}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{3x}−\mathrm{2}}\mathrm{dx}\:=\int_{\mathrm{3}} ^{\mathrm{4}} \:\frac{\mathrm{dx}}{\mathrm{x}−\mathrm{2}}=\left[\mathrm{ln}\mid\mathrm{x}−\mathrm{2}\mid\right]_{\mathrm{3}} ^{\mathrm{4}} \:=\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$

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