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Question Number 120703 by bobhans last updated on 02/Nov/20

∫ (dx/(x^2 (√(9+x^2 )))) ?

$$\:\int\:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \:\sqrt{\mathrm{9}+\mathrm{x}^{\mathrm{2}} }}\:? \\ $$

Answered by john santu last updated on 02/Nov/20

let x = 3tan r ⇒dx=3sec^2 r dr ∫ ((3 sec^2 r dr)/(9tan^2 r.3sec r)) = (1/9)∫(1/(cos r)).cot^2 r dr =(1/9)∫ ((cos r dr)/(sin^2 r)) = (1/9)∫ ((d(sin r))/(sin^2 r)) =−(1/(9sin r)) + c = −((√(9+x^2 ))/(9x)) + c

$${let}\:{x}\:=\:\mathrm{3tan}\:{r}\:\Rightarrow{dx}=\mathrm{3sec}\:^{\mathrm{2}} {r}\:{dr} \\ $$$$\int\:\frac{\mathrm{3}\:\mathrm{sec}\:^{\mathrm{2}} {r}\:{dr}}{\mathrm{9tan}\:^{\mathrm{2}} {r}.\mathrm{3sec}\:{r}}\:=\:\frac{\mathrm{1}}{\mathrm{9}}\int\frac{\mathrm{1}}{\mathrm{cos}\:{r}}.\mathrm{cot}\:^{\mathrm{2}} {r}\:{dr} \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\int\:\frac{\mathrm{cos}\:{r}\:{dr}}{\mathrm{sin}\:^{\mathrm{2}} {r}}\:=\:\frac{\mathrm{1}}{\mathrm{9}}\int\:\frac{{d}\left(\mathrm{sin}\:{r}\right)}{\mathrm{sin}\:^{\mathrm{2}} {r}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{9sin}\:{r}}\:+\:{c}\:=\:−\frac{\sqrt{\mathrm{9}+{x}^{\mathrm{2}} }}{\mathrm{9}{x}}\:+\:{c} \\ $$

Commented by malwaan last updated on 04/Nov/20

yes sir you are right

$${yes}\:{sir} \\ $$$${you}\:{are}\:{right} \\ $$

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