\( \int\displaystyle\frac{x+\displaystyle\frac{a}{x}}{x^{2}+\displaystyle\frac{a^{2}}{x^{2}}}\,dx,\ a\gt 0 \) is equal to (where C is an arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{1}{4}\log_{e}|x^{4}+a^{2}|+\displaystyle\frac{1}{2}\tan^{-1}\displaystyle\frac{x^{2}}{a}+\mathrm{C} \)
Multiplying the numerator and the denominator by \( x^{2} \):
\( \displaystyle \frac{x+\frac{a}{x}}{x^{2}+\frac{a^{2}}{x^{2}}}=\frac{x^{3}+ax}{x^{4}+a^{2}} \)
So the integral splits into two parts:
\( \displaystyle I=\int\frac{x^{3}}{x^{4}+a^{2}}\,dx+a\int\frac{x}{x^{4}+a^{2}}\,dx \)
First part: the numerator is \( \frac{1}{4} \) of the derivative of the denominator.
\( \displaystyle \int\frac{x^{3}}{x^{4}+a^{2}}\,dx=\frac{1}{4}\log_{e}|x^{4}+a^{2}| \)
Second part: put \( x^{2}=t \), so \( x\,dx=\frac{1}{2}dt \).
\( \displaystyle a\int\frac{x}{x^{4}+a^{2}}\,dx=\frac{a}{2}\int\frac{dt}{t^{2}+a^{2}} \)
\( \displaystyle =\frac{a}{2}\cdot\frac{1}{a}\tan^{-1}\frac{t}{a} \)
\( \displaystyle =\frac{1}{2}\tan^{-1}\frac{x^{2}}{a} \)
Hence
\( \displaystyle I=\frac{1}{4}\log_{e}|x^{4}+a^{2}|+\frac{1}{2}\tan^{-1}\frac{x^{2}}{a}+C \)
Why the other options are incorrect
Option (2) \( \displaystyle\frac{1}{2}\log_{e}|x^{2}+a|+\displaystyle\frac{1}{4}\tan^{-1}\displaystyle\frac{x^{4}}{a^{2}}+\mathrm{C} \) — Incorrect. The simplified denominator is \( x^{4}+a^{2} \), so the log term must be \( \log_{e}|x^{4}+a^{2}| \), not \( \log_{e}|x^{2}+a| \).
Option (3) \( \displaystyle\frac{1}{4}\log_{e}|x^{4}+a^{2}|+\displaystyle\frac{a}{2}\tan^{-1}\displaystyle\frac{x^{2}}{a}+\mathrm{C} \) — Incorrect. The factor \( \frac{1}{a} \) in \( \int\frac{dt}{t^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{t}{a} \) has been left out, so the coefficient \( \frac{a}{2} \) should be \( \frac{1}{2} \).
Option (4) \( \displaystyle\frac{1}{2}\log_{e}|x^{2}+a|+\displaystyle\frac{a}{4}\tan^{-1}\displaystyle\frac{x^{4}}{a^{2}}+\mathrm{C} \) — Incorrect. The log term must be \( \log_{e}|x^{4}+a^{2}| \), not \( \log_{e}|x^{2}+a| \).