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Question 17 of 39

Q.∫1x2−9 dx=\int \dfrac{1}{\sqrt{x^2 - 9}}\, dx = ______.

(a) 13log⁡∣x+x2−9∣+c\dfrac{1}{3} \log |x + \sqrt{x^2 - 9}| + c
(b) log⁡∣x+x2−9∣+c\log |x + \sqrt{x^2 - 9}| + c
(c) 3log⁡∣x+x2−9∣+c3\log |x + \sqrt{x^2 - 9}| + c
(d) log⁡∣x−x2−9∣+c\log |x - \sqrt{x^2 - 9}| + c
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2022MCQ· 1mImportance★★★★★
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Using the standard result ∫dxx2−a2=log⁡∣x+x2−a2∣+c\int \dfrac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + c with a2=9a^2 = 9, the answer is log⁡∣x+x2−9∣+c\log\left|x + \sqrt{x^2 - 9}\right| + c.

One of the standard forms of integration is

∫dxx2−a2=log⁡∣x+x2−a2∣+c.\int \frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + c.

Here the integrand is 1x2−9\dfrac{1}{\sqrt{x^2 - 9}}, so a2=9a^2 = 9, giving …

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