Q.23 Prove that : ∫ 1/a² - x² dx = 1/2alog(a + x/a - x) + c .

Answer: Split 1/a² - x² into partial fractions and integrate. 1/a² - x² = 1/(a + x)(a - x) = 1/2a[(a - x) + (a + x)/(a + x)(a - x)] = 1/2a[1/a + x + 1/a - x] . ∫ 1/a² - x²dx = 1/2a[∫ 1/a + xdx + ∫ 1/a - xdx] = 1/2a[log|a + x| - log|a - x|] + c = 1/2alog|a + x/a - x| + c . Hence proved.

Prove that : \(\int \frac{1}{a^2 - x^2}\,dx = \frac{1}{2a}\log\left(\frac{a + x}{a - x}\right) + c\).

Q.23 Prove that : \(\int \frac{1}{a^2 - x^2}\,dx = \frac{1}{2a}\log\left(\frac{a + x}{a - x}\right) + c\).

Doubtly Asked 29 Oct 2026 Board exam · March 2024 hsc 12th Save WhatsApp Report

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Answer: Split \(\frac{1}{a^2 - x^2}\) into partial fractions and integrate.

\(\frac{1}{a^2 - x^2} = \frac{1}{(a + x)(a - x)} = \frac{1}{2a}\left[\frac{(a - x) + (a + x)}{(a + x)(a - x)}\right] = \frac{1}{2a}\left[\frac{1}{a + x} + \frac{1}{a - x}\right]\).

\(\int \frac{1}{a^2 - x^2}dx = \frac{1}{2a}\left[\int \frac{1}{a + x}dx + \int \frac{1}{a - x}dx\right] = \frac{1}{2a}\left[\log|a + x| - \log|a - x|\right] + c\)

\(= \frac{1}{2a}\log\left|\frac{a + x}{a - x}\right| + c\). Hence proved.

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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Prove that : \(\int \frac{1}{a^2 - x^2}\,dx = \frac{1}{2a}\log\left(\frac{a + x}{a - x}\right) + c\).

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