Question

$\int \frac{x^2}{(x^2-1)(x^2+1)} dx =$

• A. $\frac{1}{4}\log\left|\frac{x+1}{x-1}\right| - \frac{1}{2}\tan^{-1}x + c$
• B. $\frac{1}{4}\log\left|\frac{x-1}{x+1}\right| + \frac{1}{2}\tan^{-1}x + c$
• C. $\frac{1}{4}\log\left|\frac{x-1}{x+1}\right| - \frac{1}{2}\tan^{-1}x + c$
• D. $\frac{1}{4}\log\left|\frac{x+1}{x-1}\right| + \frac{1}{2}\tan^{-1}x + c$

Hint:

Integration using Partial Fractions.

• Decompose rational expressions into simpler fractions.
• Integrate each part using standard results.
• Watch out for even powers in denominators — often lead to $\tan^-1$.

Answer 1

The Correct Option is B

Split the integrand into partial fractions: \[ \frac{x^2}{(x^2-1)(x^2+1)} = \frac{1}{2(x^2-1)} + \frac{1}{2(x^2+1)} \] Further, decompose: \[ \frac{1}{x^2-1} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right) \] So: \[ \int \frac{x^2}{(x^2-1)(x^2+1)} dx = \frac{1}{4}\ln\left|\frac{x-1}{x+1}\right| + \frac{1}{2}\tan^{-1}x + c \]

Answer 2

Step 1: Analyze the integral
We need to evaluate:
\[ \int \frac{x^2}{(x^2 - 1)(x^2 + 1)} \, dx \]

Step 2: Simplify the integrand using partial fractions
Express the integrand as:
\[ \frac{x^2}{(x^2 - 1)(x^2 + 1)} = \frac{x^2}{(x-1)(x+1)(x^2 + 1)} \]
We write partial fractions in the form:
\[ \frac{x^2}{(x-1)(x+1)(x^2 + 1)} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx + D}{x^2 + 1} \]

Step 3: Multiply both sides by the denominator
\[ x^2 = A(x+1)(x^2+1) + B(x-1)(x^2+1) + (Cx + D)(x^2 - 1) \]

Step 4: Expand and collect terms
Expand each term:
- \(A(x+1)(x^2+1) = A(x^3 + x^2 + x + 1)\)
- \(B(x-1)(x^2+1) = B(x^3 - x^2 + x - 1)\)
- \((Cx + D)(x^2 - 1) = Cx^3 - Cx + Dx^2 - D\)

Summing up:
\[ x^2 = (A + B + C) x^3 + (A - B + D) x^2 + (A + B - C) x + (A - B - D) \]

Step 5: Equate coefficients from both sides
Comparing with \(x^2\), coefficients on left side:
- \(x^3\) term: 0
- \(x^2\) term: 1
- \(x\) term: 0
- Constant term: 0

So, set equations:
\[ \begin{cases} A + B + C = 0 \\ A - B + D = 1 \\ A + B - C = 0 \\ A - B - D = 0 \end{cases} \]

Step 6: Solve the system
From last two equations:
\[ A - B + D = 1, \quad A - B - D = 0 \]
Adding: \(2(A - B) = 1 \implies A - B = \frac{1}{2}\)
Subtracting: \(2D = 1 \implies D = \frac{1}{2}\)

From first and third equations:
\[ A + B + C = 0, \quad A + B - C = 0 \]
Adding: \(2(A + B) = 0 \implies A + B = 0\)
Subtracting: \(2C = 0 \implies C = 0\)

From \(A + B = 0\) and \(A - B = \frac{1}{2}\):
Adding: \(2A = \frac{1}{2} \implies A = \frac{1}{4}\)
Then \(B = -A = -\frac{1}{4}\)

Step 7: Rewrite integral using partial fractions
\[ \int \left(\frac{1/4}{x-1} - \frac{1/4}{x+1} + \frac{1/2}{x^2 + 1}\right) dx \]

Step 8: Integrate term-by-term
\[ = \frac{1}{4} \int \frac{dx}{x-1} - \frac{1}{4} \int \frac{dx}{x+1} + \frac{1}{2} \int \frac{dx}{x^2 + 1} \]

Integrals are:
\[ \int \frac{dx}{x-a} = \log|x - a| + c \]
\[ \int \frac{dx}{x^2 + 1} = \tan^{-1} x + c \]

So,
\[ = \frac{1}{4} \log|x-1| - \frac{1}{4} \log|x+1| + \frac{1}{2} \tan^{-1} x + c \]
\[ = \frac{1}{4} \log \left|\frac{x-1}{x+1}\right| + \frac{1}{2} \tan^{-1} x + c \]

Final answer:
\[ \boxed{\frac{1}{4} \log \left|\frac{x-1}{x+1}\right| + \frac{1}{2} \tan^{-1} x + c} \]