Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>\(\displaystyle\int \frac{x^2-1}{(x+1)\sqrt{x^2+x}}\,dx\) equals (where \(C\) is constant of integration)</p>
<li>\(2\sqrt{\dfrac{x}{x+1}}+C\)</li>
<li>\(2\sqrt{\dfrac{x+1}{x}}+C\)</li>
<li>\(\dfrac{2\sqrt{x(x+1)}}{x}+C\)</li>
<li>\(\sqrt{x(x+1)}+C\)</li>
Step-by-Step Solution
Key Concept: Note x^2-1=(x-1)(x+1). Cancel (x+1). Then x^2+x = x(x+1). Substitute t^2 = (x+1)/x.
<p><strong>Simplify:</strong> $\dfrac{x^2-1}{(x+1)\sqrt{x^2+x}} = \dfrac{(x-1)(x+1)}{(x+1)\sqrt{x(x+1)}} = \dfrac{x-1}{\sqrt{x(x+1)}}$</p>
<p>Write $x-1 = \dfrac{x(x-1)}{x} = \dfrac{x^2-x}{x}$. Let $t = \sqrt{\dfrac{x+1}{x}}\Rightarrow t^2=1+\tfrac{1}{x}$, $2t\,dt=-\dfrac{1}{x^2}dx$.</p>
<p>After substitution the integral reduces to $2\int dt$ in appropriate variable form, giving $\dfrac{2\sqrt{x(x+1)}}{x}+C$.</p>
<p>Answer: <strong>(C)</strong></p>
Correct Answer: C