Q.1 (A)(v) displaystyle∫ dx/((x-8)(x+7)) = _____. A. 1/15log|(x+2)/(x+1)| + c B. 1/15log|(x+8)/(x+7)| + c C. 1/15log|(x-8)/(x+7)| + c D. (x-8)(x+7) + c

Answer: (c) 1/15log|(x-8)/(x+7)| + c By partial fractions, 1/((x-8)(x+7)) = 1/15[1/(x-8) - 1/(x+7)] (since (x+7) - (x-8) = 15 ). Integrating, 1/15[log|x-8| - log|x+7|] + c = 1/15log|(x-8)/(x+7)| + c.

Chapter: Integration· Mathematics & Statistics (Commerce) Part 1

\(\displaystyle\int \frac{dx}{(x-8)(x+7)} =\) _____.

Q.1 (A)(v) \(\displaystyle\int \frac{dx}{(x-8)(x+7)} =\) _____.

A. \(\dfrac{1}{15}\log\left|\dfrac{x+2}{x+1}\right| + c\)

B. \(\dfrac{1}{15}\log\left|\dfrac{x+8}{x+7}\right| + c\)

C. \(\dfrac{1}{15}\log\left|\dfrac{x-8}{x+7}\right| + c\)

D. \((x-8)(x+7) + c\)

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Answer: (c) \(\dfrac{1}{15}\log\left|\dfrac{x-8}{x+7}\right| + c\)

By partial fractions, \(\dfrac{1}{(x-8)(x+7)} = \dfrac{1}{15}\left[\dfrac{1}{x-8} - \dfrac{1}{x+7}\right]\) (since \((x+7) - (x-8) = 15\)). Integrating, \(\dfrac{1}{15}\left[\log|x-8| - \log|x+7|\right] + c = \dfrac{1}{15}\log\left|\dfrac{x-8}{x+7}\right| + c\).

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\(\displaystyle\int \frac{dx}{(x-8)(x+7)} =\) _____.

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