\(\int\displaystyle\frac{x^3}{x^4 + 3x^2 + 2}\,dx\) is equal to (where C is a constant of integration)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\log(x^2 + 2) - \frac{1}{2}\log(x^2 + 1) + C\)
Let \(t = x^2\). Then \(dt = 2x\,dx\), so \(x^3\,dx = x^2 \cdot x\,dx = \frac{1}{2}t\,dt\).
\( \displaystyle \int \frac{x^3}{x^4 + 3x^2 + 2}\,dx = \frac{1}{2}\int \frac{t}{(t + 1)(t + 2)}\,dt \)
In partial fractions:
\( \displaystyle \frac{t}{(t + 1)(t + 2)} = \frac{A}{t + 1} + \frac{B}{t + 2} \)
Putting \(t = -1\): \(-1 = A(1)\), so \(A = -1\).
Putting \(t = -2\): \(-2 = B(-1)\), so \(B = 2\).
\( \displaystyle \frac{1}{2}\int\left(\frac{2}{t + 2} - \frac{1}{t + 1}\right)dt = \log(t + 2) - \frac{1}{2}\log(t + 1) + C \)
\( \displaystyle = \log(x^2 + 2) - \frac{1}{2}\log(x^2 + 1) + C \)
Why the other options are incorrect
Option (1) \(2\log(x^2 + 2) - \log(x^2 + 1) + C\) — Incorrect. This is twice the answer: the \(\frac{1}{2}\) from \(x\,dx = \frac{1}{2}\,dt\) is left out.
Option (2) \(\displaystyle\frac{1}{2}\log(x^2 + 2) - \log(x^2 + 1) + C\) — Incorrect. The sizes of A and B are interchanged.
Option (4) \(\displaystyle\frac{1}{4}\log(x^4 + 3x^2 + 2) + C\) — Incorrect. Its derivative is \(\frac{4x^3 + 6x}{4(x^4 + 3x^2 + 2)}\); the numerator \(x^3\) is not a multiple of \(4x^3 + 6x\), the derivative of the denominator.