Question

$\int \frac{2 x^2-1}{x^4-x^2-20} d x=$

• A. $\frac{1}{5} \log \left|\frac{x+5}{x-5}\right| + \tan^{-1}\left(\frac{x}{2}\right) + c$
• B. $\frac{1}{2\sqrt{5}} \log \left|\frac{x+\sqrt{5}}{x-\sqrt{5}}\right| + \tan^{-1}\left(\frac{x}{2}\right) + c$
• C. $\frac{1}{2\sqrt{5}} \log \left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right| + \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + c$
• D. $\frac{1}{2} \log \left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right| + \frac{1}{2}\tan^{-1}\left(\frac{x}{2}\right) + c$

Hint:

When every $x$ term in an integral is raised to an even power, substituting $x^2=t$ algebraically (without changing $dx$) makes finding the partial fractions incredibly simple. Just don't forget to swap $x^2$ back in before integrating!

Answer 1

The Correct Option is C

Step 1: Study the integral.
We need $\int\frac{2x^2-1}{x^4-x^2-20}\,dx$. Both top and bottom contain only even powers of $x$, which is a strong hint to set $x^2=t$.

Step 2: Factor the bottom in terms of $t$.
With $x^2=t$, the denominator is $t^2-t-20=(t-5)(t+4)$, so the fraction is $\frac{2t-1}{(t-5)(t+4)}$.

Step 3: Split using partial fractions.
Write $\frac{2t-1}{(t-5)(t+4)}=\frac{A}{t-5}+\frac{B}{t+4}$, so $2t-1=A(t+4)+B(t-5)$.

Step 4: Find $A$ and $B$.
Put $t=5$: $9=9A\Rightarrow A=1$. Put $t=-4$: $-9=-9B\Rightarrow B=1$. So the split is $\frac{1}{t-5}+\frac{1}{t+4}$.

Step 5: Go back to $x$.
Replace $t$ by $x^2$: \[ I=\int\left(\frac{1}{x^2-5}+\frac{1}{x^2+4}\right)dx. \]

Step 6: Use the two standard formulas.
With $\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\log\left|\frac{x-a}{x+a}\right|$ (here $a=\sqrt5$) and $\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}$ (here $a=2$): \[ I=\frac{1}{2\sqrt5}\log\left|\frac{x-\sqrt5}{x+\sqrt5}\right|+\frac{1}{2}\tan^{-1}\frac{x}{2}+c. \]

Step 7: Match the option.
This is option (3).
\[ \boxed{\frac{1}{2\sqrt5}\log\left|\frac{x-\sqrt5}{x+\sqrt5}\right|+\frac{1}{2}\tan^{-1}\frac{x}{2}+c} \]