Integral Calculus
Substitution to rationalize surd denominator
MMTS_Full_Test_02
Grade 12
Question:
If $I=\displaystyle\int\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}\,dx$, then $I$ equals
(A) $\dfrac{\sqrt{2x^4-2x^2+1}}{x^2}+C$
(B) $\dfrac{\sqrt{2x^4-2x^2+1}}{x}+C$
(C) $\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$
(D) None of these
Step-by-Step Solution
Key Concept: Divide numerator and denominator by $x^4$. Let $t=2-2/x^2+1/x^4$; then $dt=4(1/x^3-1/x^5)dx$, matching the numerator $(1/x^3-1/x^5)dx=dt/4$. Integral becomes $\frac{1}{4}\int t^{-1/2}dt=\frac{1}{2}\sqrt{t}+C$.
$I=\dfrac{1}{2}\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+C=\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$.
Correct Answer: (C) $\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+C$