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Question Number 96864 by bemath last updated on 05/Jun/20

∫ (dy/(y^2 (5−y^2 ))) ?

$$\int\:\frac{\mathrm{dy}}{\mathrm{y}^{\mathrm{2}} \left(\mathrm{5}−\mathrm{y}^{\mathrm{2}} \right)}\:? \\ $$

Commented by bemath last updated on 05/Jun/20

thank you both

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$

Answered by Sourav mridha last updated on 05/Jun/20

=(1/5)[∫(dy/y^2 ) +∫(dy/(((√5))^2 −y^2 ))] =−(1/(5y))+(1/(10(√5)))ln[(((√5)+y)/((√5)−y))]+c

$$=\frac{\mathrm{1}}{\mathrm{5}}\left[\int\frac{\boldsymbol{{dy}}}{\boldsymbol{{y}}^{\mathrm{2}} }\:+\int\frac{\boldsymbol{{dy}}}{\left(\sqrt{\mathrm{5}}\right)^{\mathrm{2}} −\boldsymbol{{y}}^{\mathrm{2}} }\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}\boldsymbol{{y}}}+\frac{\mathrm{1}}{\mathrm{10}\sqrt{\mathrm{5}}}\boldsymbol{{ln}}\left[\frac{\sqrt{\mathrm{5}}+\boldsymbol{{y}}}{\sqrt{\mathrm{5}}−\boldsymbol{{y}}}\right]+\boldsymbol{{c}} \\ $$

Answered by mathmax by abdo last updated on 05/Jun/20

I =∫ (dx/(x^2 (5−x^2 ))) let decompose F(x) =(1/(x^2 (5−x^2 ))) =(1/(x^2 ((√5)−x)((√5)+x))) F(x) =(a/x) +(b/x^2 ) +(c/((√5)−x)) +(d/((√5)+x)) b=(1/5) , c =(1/(5(2(√5)))) =(1/(10(√5))) ,d =(1/(5(2(√5)))) =(1/(10(√5))) ⇒ F(x)=(a/x) +(1/(5x^2 )) +(1/(10(√5)((√5)−x))) +(1/(10(√5)((√5)+x))) lim_(x→+∞) xF(x) =0 =a−c +d ⇒a =c−d =0 ⇒ F(x) =(1/(5x^2 )) +(1/(10(√5)((√5)−x))) +(1/(10(√5)((√5)+x))) ⇒ I =−(1/(5x))−(1/(10(√5)))ln∣x−(√5)∣+(1/(10(√5)))ln∣x+(√5)∣ +C =−(1/(5x)) +(1/(10(√5)))ln∣((x+(√5))/(x−(√5)))∣ +C

$$\mathrm{I}\:=\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{5}−\mathrm{x}^{\mathrm{2}} \right)}\:\:\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \left(\mathrm{5}−\mathrm{x}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \left(\sqrt{\mathrm{5}}−\mathrm{x}\right)\left(\sqrt{\mathrm{5}}+\mathrm{x}\right)} \\ $$$$\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{a}}{\mathrm{x}}\:+\frac{\mathrm{b}}{\mathrm{x}^{\mathrm{2}} }\:+\frac{\mathrm{c}}{\sqrt{\mathrm{5}}−\mathrm{x}}\:+\frac{\mathrm{d}}{\sqrt{\mathrm{5}}+\mathrm{x}} \\ $$$$\mathrm{b}=\frac{\mathrm{1}}{\mathrm{5}}\:,\:\:\:\mathrm{c}\:=\frac{\mathrm{1}}{\mathrm{5}\left(\mathrm{2}\sqrt{\mathrm{5}}\right)}\:=\frac{\mathrm{1}}{\mathrm{10}\sqrt{\mathrm{5}}}\:,\mathrm{d}\:=\frac{\mathrm{1}}{\mathrm{5}\left(\mathrm{2}\sqrt{\mathrm{5}}\right)}\:=\frac{\mathrm{1}}{\mathrm{10}\sqrt{\mathrm{5}}}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{\mathrm{x}}\:+\frac{\mathrm{1}}{\mathrm{5x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{10}\sqrt{\mathrm{5}}\left(\sqrt{\mathrm{5}}−\mathrm{x}\right)}\:+\frac{\mathrm{1}}{\mathrm{10}\sqrt{\mathrm{5}}\left(\sqrt{\mathrm{5}}+\mathrm{x}\right)} \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow+\infty} \mathrm{xF}\left(\mathrm{x}\right)\:=\mathrm{0}\:=\mathrm{a}−\mathrm{c}\:+\mathrm{d}\:\Rightarrow\mathrm{a}\:=\mathrm{c}−\mathrm{d}\:=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{F}\left(\mathrm{x}\right)\:=\frac{\mathrm{1}}{\mathrm{5x}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{10}\sqrt{\mathrm{5}}\left(\sqrt{\mathrm{5}}−\mathrm{x}\right)}\:+\frac{\mathrm{1}}{\mathrm{10}\sqrt{\mathrm{5}}\left(\sqrt{\mathrm{5}}+\mathrm{x}\right)}\:\Rightarrow \\ $$$$\mathrm{I}\:=−\frac{\mathrm{1}}{\mathrm{5x}}−\frac{\mathrm{1}}{\mathrm{10}\sqrt{\mathrm{5}}}\mathrm{ln}\mid\mathrm{x}−\sqrt{\mathrm{5}}\mid+\frac{\mathrm{1}}{\mathrm{10}\sqrt{\mathrm{5}}}\mathrm{ln}\mid\mathrm{x}+\sqrt{\mathrm{5}}\mid\:+\mathrm{C} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5x}}\:+\frac{\mathrm{1}}{\mathrm{10}\sqrt{\mathrm{5}}}\mathrm{ln}\mid\frac{\mathrm{x}+\sqrt{\mathrm{5}}}{\mathrm{x}−\sqrt{\mathrm{5}}}\mid\:+\mathrm{C} \\ $$

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