Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>The value of \(\displaystyle\int\frac{x^n-1}{x^{n+1}\sqrt{x^{2n}+x^n+1}}\,dx\) is equal to</p>
<li>\(\dfrac{1}{n}\sqrt{x^{-n}+1+x^n}+C\)</li>
<li>\(\dfrac{2}{n}\sqrt{\dfrac{1}{x^n}+1+x^n}+C\)</li>
<li>\(\dfrac{1}{n}\tan^{-1}\!\left(\sqrt{x^{-n}+1+x^n}\right)+C\)</li>
<li>\(-\dfrac{2}{n}\sqrt{x^{-2n}+x^{-n}+1}+C\)</li>
Step-by-Step Solution
Key Concept: Divide numerator and denominator by x^(n+1). Let t = x^(-n)+1+x^n, then dt = n(x^(n-1)-x^(-n-1))dx.
<p>Divide top and bottom by $x^{n+1}$:</p>
<p>$$\int\frac{x^{-2}-x^{-(n+2)}}{\sqrt{x^{-2n}+x^{-n}+1}}\cdot\ldots$$</p>
<p>More carefully: multiply top by $x^{-n}$, bottom by $x^n$:</p>
<p>$$\int\frac{x^{-1}-x^{-n-1}}{\sqrt{x^{-2n}+x^{-n}+1}}\,dx\cdot\frac{1}{n}\cdot n$$</p>
<p>Let $t^2 = x^{-2n}+x^{-n}+1\Rightarrow 2t\,dt = (-2nx^{-2n-1}-nx^{-n-1})\,dx$.</p>
<p>Simplify to get $\displaystyle\frac{-1}{n}\int\frac{dt}{\sqrt{t^2}}\cdot\ldots = -\frac{2}{n}\sqrt{x^{-2n}+x^{-n}+1}+C$. Answer: <strong>ABD</strong></p>
Correct Answer: ABD