A standard antiderivative is dx x^ 2 -a^ 2 = 1 2a | x-a x+a |+C

MATHEMATICS · Multi-part · CBSE · Mathematics · Integrals · Integrals

Problem

A standard antiderivative is ∫dxx2−a2=12aln⁡∣x−ax+a∣+C,\int \frac{dx}{x^{2}-a^{2}}=\frac{1}{2a}\ln\Bigl|\frac{x-a}{x+a}\Bigr|+C, where aa is a non-zero constant. Answer the following questions. (i) Explain why the formula is stated for a≠0a\neq 0 and x≠±ax\neq \pm a. What would go wrong if a=0a=0 or x=±ax=\pm a? (ii) Show algebraically that replacing aa by ∣a∣|a| leaves the right-hand side unchanged. Why does this confirm that one may assume a>0a>0 without loss of generality? (iii) Taking a>0a>0, rewrite the antiderivative so that only the absolute value ∣x∣|x| appears inside the logarithm. State clearly the domain on which your rewritten form is valid.

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