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Question Number 128900 by bemath last updated on 11/Jan/21
∫ ((4x+5)/((x+2)(x+3)(x+4)(x+5)+1)) dx?
$$\:\int\:\frac{\mathrm{4x}+\mathrm{5}}{\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{4}\right)\left(\mathrm{x}+\mathrm{5}\right)+\mathrm{1}}\:\mathrm{dx}?\: \\ $$
Answered by liberty last updated on 11/Jan/21
∫ ((4x+5)/((x^2 +7x+10)(x^2 +7x+12)+1)) dx=(∗) (x^2 +7x+10)(x^2 +7x+12)+1 let x^2 +7x=u ⇒(u+10)(u+12)+1=u^2 +22u+121 = (u+11)^2 =(x^2 +7x+11)^2 (∗) ∫ ((4x+5)/((x^2 +7x+11)^2 )) dx = ∫ ((2(2x+7)−9)/((x^2 +7x+11)^2 )) dx = ∫((2(2x+7))/((x^2 +7x+11)^2 )) dx−∫(9/((x^2 +7x+11)^2 )) dx = −(2/(x^2 +7x+11)) −∫ (9/((x^2 +7x+11)^2 )) dx next..
$$\:\int\:\frac{\mathrm{4x}+\mathrm{5}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{10}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{12}\right)+\mathrm{1}}\:\mathrm{dx}=\left(\ast\right) \\ $$$$\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{10}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{12}\right)+\mathrm{1}\: \\ $$$$\mathrm{let}\:\mathrm{x}^{\mathrm{2}} +\mathrm{7x}=\mathrm{u}\:\Rightarrow\left(\mathrm{u}+\mathrm{10}\right)\left(\mathrm{u}+\mathrm{12}\right)+\mathrm{1}=\mathrm{u}^{\mathrm{2}} +\mathrm{22u}+\mathrm{121} \\ $$$$\:=\:\left(\mathrm{u}+\mathrm{11}\right)^{\mathrm{2}} =\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{11}\right)^{\mathrm{2}} \\ $$$$\left(\ast\right)\:\int\:\frac{\mathrm{4x}+\mathrm{5}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{11}\right)^{\mathrm{2}} }\:\mathrm{dx}\:= \\ $$$$\:\int\:\frac{\mathrm{2}\left(\mathrm{2x}+\mathrm{7}\right)−\mathrm{9}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{11}\right)^{\mathrm{2}} }\:\mathrm{dx}\:=\:\int\frac{\mathrm{2}\left(\mathrm{2x}+\mathrm{7}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{11}\right)^{\mathrm{2}} }\:\mathrm{dx}−\int\frac{\mathrm{9}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{11}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$=\:−\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{11}}\:−\int\:\frac{\mathrm{9}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{7x}+\mathrm{11}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\mathrm{next}.. \\ $$
Commented by bramlexs22 last updated on 11/Jan/21
I_2 =∫ ((9 dx)/(((x+(7/2))^2 −(((√5)/2))^2 )^2 )) let x+(7/2) = ((√5)/2) sec t I_2 =∫ ((((9(√5))/2) sec t tan t)/(((25)/4)tan^4 t)) dt I_2 =((18(√5))/(25)) ∫((cos^2 t)/(sin^3 t)) dt I_2 = ((18(√5))/(25)) (∫ csc^3 tdt −∫ csc t dt )
$$\mathrm{I}_{\mathrm{2}} =\int\:\frac{\mathrm{9}\:\mathrm{dx}}{\left(\left(\mathrm{x}+\frac{\mathrm{7}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\mathrm{let}\:\mathrm{x}+\frac{\mathrm{7}}{\mathrm{2}}\:=\:\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:\mathrm{sec}\:\mathrm{t} \\ $$$$\mathrm{I}_{\mathrm{2}} =\int\:\frac{\frac{\mathrm{9}\sqrt{\mathrm{5}}}{\mathrm{2}}\:\mathrm{sec}\:\mathrm{t}\:\mathrm{tan}\:\mathrm{t}}{\frac{\mathrm{25}}{\mathrm{4}}\mathrm{tan}\:^{\mathrm{4}} \mathrm{t}}\:\mathrm{dt} \\ $$$$\mathrm{I}_{\mathrm{2}} =\frac{\mathrm{18}\sqrt{\mathrm{5}}}{\mathrm{25}}\:\int\frac{\mathrm{cos}\:^{\mathrm{2}} \mathrm{t}}{\mathrm{sin}\:^{\mathrm{3}} \mathrm{t}}\:\mathrm{dt} \\ $$$$\mathrm{I}_{\mathrm{2}} =\:\frac{\mathrm{18}\sqrt{\mathrm{5}}}{\mathrm{25}}\:\left(\int\:\mathrm{csc}^{\mathrm{3}} \:\mathrm{tdt}\:−\int\:\mathrm{csc}\:\mathrm{t}\:\mathrm{dt}\:\right) \\ $$$$ \\ $$
Answered by Olaf last updated on 11/Jan/21
F(x) = ∫((4x+5)/((x+2)(x+3)(x+4)(x+5)+1))dx Let u = x+(7/2) F(x) = ∫((4u−9)/((u−(3/2))(u−(1/2))(u+(1/2))(u+(3/2))+1))du F(x) = ∫((4u−9)/((u^2 −(1/4))(u^2 −(9/4))+1))du F(x) = ∫((4u−9)/(u^4 −(5/2)u^2 +((25)/(16))))du F(x) = ∫((4u−9)/((u^2 −(5/4))^2 ))du F(x) = ∫[(4/(u−(5/4)))−(4/((u^2 −(5/4))^2 ))]du F(x) = 4ln∣u−(5/4)∣+((8u)/(5(u^2 −(5/4))))−((16)/( (√5)))atanh(((2u)/( (√5))))+C F(x) = 4ln∣x+(9/4)∣+((4(x+7))/(5(x^2 +7x+11))) −((16)/( (√5)))atanh(((2x+7)/( (√5))))+C
$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{4}{x}+\mathrm{5}}{\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}+\mathrm{4}\right)\left({x}+\mathrm{5}\right)+\mathrm{1}}{dx} \\ $$$$\mathrm{Let}\:{u}\:=\:{x}+\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{4}{u}−\mathrm{9}}{\left({u}−\frac{\mathrm{3}}{\mathrm{2}}\right)\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({u}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({u}+\frac{\mathrm{3}}{\mathrm{2}}\right)+\mathrm{1}}{du} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{4}{u}−\mathrm{9}}{\left({u}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)\left({u}^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}\right)+\mathrm{1}}{du} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{4}{u}−\mathrm{9}}{{u}^{\mathrm{4}} −\frac{\mathrm{5}}{\mathrm{2}}{u}^{\mathrm{2}} +\frac{\mathrm{25}}{\mathrm{16}}}{du} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\frac{\mathrm{4}{u}−\mathrm{9}}{\left({u}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} }{du} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\left[\frac{\mathrm{4}}{{u}−\frac{\mathrm{5}}{\mathrm{4}}}−\frac{\mathrm{4}}{\left({u}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} }\right]{du} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{4ln}\mid{u}−\frac{\mathrm{5}}{\mathrm{4}}\mid+\frac{\mathrm{8}{u}}{\mathrm{5}\left({u}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right)}−\frac{\mathrm{16}}{\:\sqrt{\mathrm{5}}}\mathrm{atanh}\left(\frac{\mathrm{2}{u}}{\:\sqrt{\mathrm{5}}}\right)+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{4ln}\mid{x}+\frac{\mathrm{9}}{\mathrm{4}}\mid+\frac{\mathrm{4}\left({x}+\mathrm{7}\right)}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{11}\right)} \\ $$$$−\frac{\mathrm{16}}{\:\sqrt{\mathrm{5}}}\mathrm{atanh}\left(\frac{\mathrm{2}{x}+\mathrm{7}}{\:\sqrt{\mathrm{5}}}\right)+\mathrm{C} \\ $$
Answered by MJS_new last updated on 11/Jan/21
∫((4x+5)/((x+2)(x+3)(x+4)(x+5)+1))dx= =∫((4x+5)/((x^2 +7x+11)^2 ))dx= [Ostrogradski] =((18x+53)/(5(x^2 +7x+11)))+((18)/5)∫(dx/(x^2 +7x+11))= =((18x+53)/(5(x^2 +7x+11)))+((36(√5))/(25))∫((1/(2x+7−(√5)))−(1/(2x+7+(√5))))dx= =((18x+53)/(5(x^2 +7x+11)))+((18(√5))/(25))(ln ∣2x+7−(√5)∣ −ln ∣2x+7+(√5)∣ = =((18x+53)/(5(x^2 +7x+11)))+((18(√5))/(25))ln ∣((2x+7−(√5))/(2x+7+(√5)))∣ +C
$$\int\frac{\mathrm{4}{x}+\mathrm{5}}{\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}+\mathrm{4}\right)\left({x}+\mathrm{5}\right)+\mathrm{1}}{dx}= \\ $$$$=\int\frac{\mathrm{4}{x}+\mathrm{5}}{\left({x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{11}\right)^{\mathrm{2}} }{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}\right] \\ $$$$=\frac{\mathrm{18}{x}+\mathrm{53}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{11}\right)}+\frac{\mathrm{18}}{\mathrm{5}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{11}}= \\ $$$$=\frac{\mathrm{18}{x}+\mathrm{53}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{11}\right)}+\frac{\mathrm{36}\sqrt{\mathrm{5}}}{\mathrm{25}}\int\left(\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{7}−\sqrt{\mathrm{5}}}−\frac{\mathrm{1}}{\mathrm{2}{x}+\mathrm{7}+\sqrt{\mathrm{5}}}\right){dx}= \\ $$$$=\frac{\mathrm{18}{x}+\mathrm{53}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{11}\right)}+\frac{\mathrm{18}\sqrt{\mathrm{5}}}{\mathrm{25}}\left(\mathrm{ln}\:\mid\mathrm{2}{x}+\mathrm{7}−\sqrt{\mathrm{5}}\mid\:−\mathrm{ln}\:\mid\mathrm{2}{x}+\mathrm{7}+\sqrt{\mathrm{5}}\mid\:=\right. \\ $$$$=\frac{\mathrm{18}{x}+\mathrm{53}}{\mathrm{5}\left({x}^{\mathrm{2}} +\mathrm{7}{x}+\mathrm{11}\right)}+\frac{\mathrm{18}\sqrt{\mathrm{5}}}{\mathrm{25}}\mathrm{ln}\:\mid\frac{\mathrm{2}{x}+\mathrm{7}−\sqrt{\mathrm{5}}}{\mathrm{2}{x}+\mathrm{7}+\sqrt{\mathrm{5}}}\mid\:+{C} \\ $$

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