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Question Number 75512 by Master last updated on 12/Dec/19

Commented by MJS last updated on 12/Dec/19

this is a standard integral; you can find it on any table of integrals

$$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{standard}\:\mathrm{integral};\:\mathrm{you}\:\mathrm{can}\:\mathrm{find}\:\mathrm{it}\:\mathrm{on} \\ $$$$\mathrm{any}\:\mathrm{table}\:\mathrm{of}\:\mathrm{integrals} \\ $$

Commented by JDamian last updated on 12/Dec/19

Tap on Study > Integral Calculus

$${Tap}\:{on}\:\boldsymbol{\mathrm{Study}}\:>\:\boldsymbol{\mathrm{Integral}}\:\boldsymbol{\mathrm{Calculus}} \\ $$

Answered by Kunal12588 last updated on 12/Dec/19

I=∫(√(x^2 −a^2 ))dx =x(√(x^2 −a^2 ))−∫(x^2 /(√(x^2 −a^2 )))dx =x(√(x^2 −a^2 ))−∫((x^2 −a^2 )/(√(x^2 −a^2 )))dx−a^2 ∫(1/(√(x^2 −a^2 )))dx =x(√(x^2 −a^2 ))−I−a^2 log∣x+(√(x^2 −a^2 ))∣dx =(x/2)(√(x^2 −a^2 ))−(a^2 /2)log∣x+(√(x^2 −a^2 ))∣dx

$${I}=\int\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }{dx} \\ $$$$={x}\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }−\int\frac{{x}^{\mathrm{2}} }{\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{dx} \\ $$$$={x}\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }−\int\frac{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{dx}−{a}^{\mathrm{2}} \int\frac{\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{dx} \\ $$$$={x}\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }−{I}−{a}^{\mathrm{2}} {log}\mid{x}+\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\mid{dx} \\ $$$$=\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{log}\mid{x}+\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\mid{dx} \\ $$

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