Indefinite Integration
Integration of Rational Functions
Grade None
Question:
<p>[JEE Main 2020] \(\displaystyle\int\frac{x+2}{\sqrt{x^2-1}}\,dx\) equals (where \(C\) is a constant)</p>
<li>\(\sqrt{x^2-1}+2\ln\!\left|x+\sqrt{x^2-1}\right|+C\)</li>
<li>\(\sqrt{x^2-1}+2\cosh^{-1}x+C\)</li>
<li>\(\sqrt{x^2-1}+2\ln\!\left|x+\sqrt{x^2-1}\right|+C\)</li>
<li>\(2\sqrt{x^2-1}+\ln\!\left|x+\sqrt{x^2-1}\right|+C\)</li>
Step-by-Step Solution
Key Concept: Split: \int(x+2)/\sqrt{x^2-1} = \intx/\sqrt{x^2-1} dx + 2\int1/\sqrt{x^2-1} dx. First = \sqrt{x^2-1}, second = cosh⁻^1x = ln|x+\sqrt{x^2-1}|.
<p><strong>Split:</strong> \(\displaystyle\int\frac{x+2}{\sqrt{x^2-1}}\,dx = \underbrace{\int\frac{x}{\sqrt{x^2-1}}\,dx}_{=\sqrt{x^2-1}}+2\underbrace{\int\frac{dx}{\sqrt{x^2-1}}}_{=\cosh^{-1}x=\ln|x+\sqrt{x^2-1}|}\)</p>
<p>\[= \sqrt{x^2-1}+2\ln\!\left|x+\sqrt{x^2-1}\right|+C\]</p>
<p>Answer: <strong>(B)</strong> (cosh⁻¹x form is equivalent)</p>
Correct Answer: B