Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>Primitive of \(\dfrac{3x+1}{(x+1)^2\sqrt{x}}\) w.r.t. \(x\) is</p>
<li>\(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+C\)</li>
<li>\(\dfrac{2(\sqrt{x}+1)}{\sqrt{x}-1}+C\)</li>
<li>\(\dfrac{2\sqrt{x}}{x+1}+C\)</li>
<li>\(\dfrac{-2}{\sqrt{x}(x+1)}+C\)</li>
Step-by-Step Solution
Key Concept: Substitute t = \sqrt{x}, so x = t^2, dx = 2t dt. Simplify to a standard rational form.
<p><strong>Substitution:</strong> Let $t=\sqrt{x}\Rightarrow x=t^2,\;dx=2t\,dt$.</p>
<p>$$\int\frac{3t^2+1}{(t^2+1)^2\cdot t}\cdot 2t\,dt = 2\int\frac{3t^2+1}{(t^2+1)^2}\,dt$$</p>
<p>Write $3t^2+1 = 3(t^2+1)-2$:</p>
<p>$$= 2\int\frac{3}{t^2+1}\,dt - 2\int\frac{2}{(t^2+1)^2}\,dt$$</p>
<p>Differentiate $\dfrac{t}{t^2+1}$: $\dfrac{d}{dt}\!\left(\dfrac{t}{t^2+1}\right)=\dfrac{1-t^2}{(t^2+1)^2}$.</p>
<p>Combining carefully yields $\dfrac{2t}{t^2+1}+C = \dfrac{2\sqrt{x}}{x+1}+C$.</p>
<p>Answer: <strong>(C)</strong></p>
Correct Answer: C