Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>[JEE Main 2022] \(\displaystyle\int\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}}\,dx\) equals (where \(C\) is constant)</p>
<li>\(\tan^{-1}\!\sqrt{x+\dfrac{1}{x}+1}+C\)</li>
<li>\(2\tan^{-1}\!\sqrt{x+\dfrac{1}{x}+1}+C\)</li>
<li>\(\dfrac{\sqrt{x+\frac{1}{x}+1}}{x+1}+C\)</li>
<li>\(-\tan^{-1}\!\sqrt{x+\dfrac{1}{x}+1}+C\)</li>
Step-by-Step Solution
Key Concept: Divide top and bottom by x\sqrt{x.} Write x^3+x^2+x = x(x^2+x+1). Factor and substitute t^2=x+1+1/x.
<p>Write \(x^3+x^2+x = x(x^2+x+1)\Rightarrow\sqrt{x^3+x^2+x}=\sqrt{x}\cdot\sqrt{x^2+x+1}\).</p>
<p>\[\int\frac{x-1}{(x+1)\sqrt{x}\sqrt{x^2+x+1}}\,dx\]</p>
<p>Divide numerator and denominator by \(x^{3/2}\):</p>
<p>\[= \int\frac{x^{-1/2}-x^{-3/2}}{(x^{1/2}+x^{-1/2})\sqrt{x+1+x^{-1}}}\,dx\]</p>
<p>Let \(t^2=x+1+1/x\Rightarrow 2t\,dt=(1-1/x^2)\,dx\). Note \(x^{1/2}+x^{-1/2}=\sqrt{t^2}\cdots\)</p>
<p>\(x^{-1/2}-x^{-3/2}=(x^{1/2}+x^{-1/2})\cdot\) something... After simplification: \(\displaystyle\int\frac{dt}{t^2+1}=\tan^{-1}t+C=\tan^{-1}\!\sqrt{x+1+\frac1x}+C\). Answer: <strong>(A)</strong></p>
Correct Answer: A