Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>\(\displaystyle\int \sqrt{\dfrac{1+\ln x}{x\ln x}}\,dx\) equals</p>
<li>\(\sqrt{\ln x\cdot(1+\ln x)}+\ln\!\left|\sqrt{\ln x}+\sqrt{1+\ln x}\right|+C\)</li>
<li>\(2\sqrt{\ln x\cdot(1+\ln x)}+C\)</li>
<li>\(\dfrac{\sqrt{1+\ln x}}{\ln x}+C\)</li>
<li>\(\ln\!\left|\sqrt{\ln x}+\sqrt{1+\ln x}\right|+C\)</li>
Step-by-Step Solution
Key Concept: Substitute u = lnx, then write \sqrt{(1+u}/u) = \sqrt{1+1/u}. Use the standard form \int\sqrt{t^2+a^2} dt.
<p><strong>Substitution:</strong> Let \(u = \ln x \Rightarrow du = \frac{dx}{x}\).</p>
<p>\[\int \sqrt{\frac{1+u}{u}}\,du = \int \sqrt{1+\frac{1}{u}}\,du\]</p>
<p>Let \(t = \sqrt{u}\), so \(u=t^2\), \(du=2t\,dt\):</p>
<p>\[= 2\int\sqrt{t^2+1}\,dt = 2\cdot\frac{t\sqrt{t^2+1}}{2}+\frac{1}{2}\ln\left|t+\sqrt{t^2+1}\right|\cdot 2 + C\]</p>
<p>\[= t\sqrt{t^2+1}+\ln\left|t+\sqrt{t^2+1}\right|+C\]</p>
<p>Back-substituting \(t=\sqrt{\ln x}\):</p>
<p>\[= \sqrt{\ln x}\cdot\sqrt{1+\ln x}+\ln\!\left|\sqrt{\ln x}+\sqrt{1+\ln x}\right|+C\]</p>
<p>Answer: <strong>(A)</strong></p>
Correct Answer: A