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

Commented by MJS last updated on 12/Dec/19

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

Commented by JDamian last updated on 12/Dec/19

$$Tapon\mathbf{Study}>\mathbf{Integral}\mathbf{Calculus}$$

Answered by Kunal12588 last updated on 12/Dec/19

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

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