Q90 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\int \dfrac{x}{x^4-16} d x=$
- A$\dfrac{1}{16} \log \left|\dfrac{x^2+4}{x^2-4}\right|+C$
- Correct answerB$\dfrac{1}{16} \log \left|\dfrac{x^2-4}{x^2+4}\right|+C$
- C$\dfrac{1}{8} \log \left|\dfrac{x^2-4}{x^2+4}\right|+C$
- D$\dfrac{1}{4} \log \left|\dfrac{x^2+4}{x^2-4}\right|+C$
Solution
Let $I = \int \dfrac{x}{x^4 - 16} dx$.
Substitute $u = x^2$, then $du = 2x dx$, which implies $x dx = \dfrac{du}{2}$.
The integral becomes $I = \int \dfrac{1}{2(u^2 - 16)} du$.
Using the partial fraction decomposition $\dfrac{1}{u^2 - 16} = \dfrac{1}{(u-4)(u+4)} = \dfrac{1}{8} \left( \dfrac{1}{u-4} - \dfrac{1}{u+4} \right)$, we have:
$I = \dfrac{1}{2} \times \dfrac{1}{8} \int \left( \dfrac{1}{u-4} - \dfrac{1}{u+4} \right) du$.
$I = \dfrac{1}{16} (\ln|u-4| - \ln|u+4|) + C$.
$I = \dfrac{1}{16} \ln \left| \dfrac{u-4}{u+4} \right| + C$.
Substituting back $u = x^2$, we get $I = \dfrac{1}{16} \ln \left| \dfrac{x^2-4}{x^2+4} \right| + C$.
Answer: $\dfrac{1}{16} \log \left|\dfrac{x^2-4}{x^2+4}\right|+C$
Substitute $u = x^2$, then $du = 2x dx$, which implies $x dx = \dfrac{du}{2}$.
The integral becomes $I = \int \dfrac{1}{2(u^2 - 16)} du$.
Using the partial fraction decomposition $\dfrac{1}{u^2 - 16} = \dfrac{1}{(u-4)(u+4)} = \dfrac{1}{8} \left( \dfrac{1}{u-4} - \dfrac{1}{u+4} \right)$, we have:
$I = \dfrac{1}{2} \times \dfrac{1}{8} \int \left( \dfrac{1}{u-4} - \dfrac{1}{u+4} \right) du$.
$I = \dfrac{1}{16} (\ln|u-4| - \ln|u+4|) + C$.
$I = \dfrac{1}{16} \ln \left| \dfrac{u-4}{u+4} \right| + C$.
Substituting back $u = x^2$, we get $I = \dfrac{1}{16} \ln \left| \dfrac{x^2-4}{x^2+4} \right| + C$.
Answer: $\dfrac{1}{16} \log \left|\dfrac{x^2-4}{x^2+4}\right|+C$
Hence, the correct answer is option: 'B'.