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0-1-dx-1-3x-1-3x-




Question Number 92344 by 675480065 last updated on 06/May/20
∫_0 ^1 (dx/( (√(1+3x))−(√(1−3x))))
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}+\mathrm{3x}}−\sqrt{\mathrm{1}−\mathrm{3x}}} \\ $$
Commented by john santu last updated on 06/May/20
∫_0 ^1  (((√(1+3x)) + (√(1−3x)))/(6x))  dx   I_1  = ∫_0 ^1  ((√(1+3x))/(6x)) dx  I_2  = ∫_0 ^1  ((√(1−3x))/(6x)) dx
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\sqrt{\mathrm{1}+\mathrm{3}{x}}\:+\:\sqrt{\mathrm{1}−\mathrm{3}{x}}}{\mathrm{6}{x}}\:\:{dx}\: \\ $$$${I}_{\mathrm{1}} \:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\sqrt{\mathrm{1}+\mathrm{3}{x}}}{\mathrm{6}{x}}\:{dx} \\ $$$${I}_{\mathrm{2}} \:=\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\sqrt{\mathrm{1}−\mathrm{3}{x}}}{\mathrm{6}{x}}\:{dx}\: \\ $$
Commented by 675480065 last updated on 06/May/20
thnks sir   i cant continue
$$\mathrm{thnks}\:\mathrm{sir}\: \\ $$$$\mathrm{i}\:\mathrm{cant}\:\mathrm{continue} \\ $$
Commented by mathmax by abdo last updated on 06/May/20
this integral is divergent   (√(1+3x))∼1+((3x)/2)  and (√(1−3x))∼1−((3x)/2)  ⇒(√(1+3x))−(√(1−3x))∼3x ⇒(1/( (√(1+3x))−(√(1−3x)))) ∼(1/(3x)) but  ∫_0 ^1  (dx/(3x))   is divergent...!
$${this}\:{integral}\:{is}\:{divergent}\:\:\:\sqrt{\mathrm{1}+\mathrm{3}{x}}\sim\mathrm{1}+\frac{\mathrm{3}{x}}{\mathrm{2}}\:\:{and}\:\sqrt{\mathrm{1}−\mathrm{3}{x}}\sim\mathrm{1}−\frac{\mathrm{3}{x}}{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{1}+\mathrm{3}{x}}−\sqrt{\mathrm{1}−\mathrm{3}{x}}\sim\mathrm{3}{x}\:\Rightarrow\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{3}{x}}−\sqrt{\mathrm{1}−\mathrm{3}{x}}}\:\sim\frac{\mathrm{1}}{\mathrm{3}{x}}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\mathrm{3}{x}}\:\:\:{is}\:{divergent}…! \\ $$

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