Question-75512

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

Commented by JDamian last updated on 12/Dec/19

T a p o n Study > Integral Calculus

Answered by Kunal12588 last updated on 12/Dec/19

I = \int \sqrt{x^{2} - a^{2}} d x

= x \sqrt{x^{2} - a^{2}} - \int \frac{x^{2}}{\sqrt{x^{2} - a^{2}}} d x

= x \sqrt{x^{2} - a^{2}} - \int \frac{x^{2} - a^{2}}{\sqrt{x^{2} - a^{2}}} d x - a^{2} \int \frac{1}{\sqrt{x^{2} - a^{2}}} d x

= x \sqrt{x^{2} - a^{2}} - I - a^{2} l o g ∣ x + \sqrt{x^{2} - a^{2}} ∣ d x

= \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} l o g ∣ x + \sqrt{x^{2} - a^{2}} ∣ d x