Question

If \(\int \frac{\log(1+x^4)}{x^3} dx = f(x) \log(\frac{1}{g(x)}) + \tan^{-1}(h(x)) + c\), then \(h(x) [f(x) + f(\frac{1}{x})] =\)

• A. \(h(x)g(-x)\)
• B. \(\frac{g(x)}{2}\)
• C. \(g(x) + g(-x)\)
• D. \(g(x)h(x)\)

Hint:

Use integration by parts and partial fractions to evaluate the integral.

Answer 1

The Correct Option is B

We are given: \[ \int \frac{\log(1 + x^4)}{x^3} \, dx = f(x) \log\left(\frac{1}{g(x)}\right) + \tan^{-1}(h(x)) + C \] We need to find the expression for: \[ h(x) [f(x) + f\left(\frac{1}{x}\right)] \] 

Step 1: Differentiating the Integral 
Let \[ I = \int \frac{\log(1 + x^4)}{x^3} \, dx \] Differentiating both sides, \[ \frac{dI}{dx} = \frac{\log(1 + x^4)}{x^3} \] Now let \( u = 1 + x^4 \implies du = 4x^3 dx \) \[ \frac{1}{x^3} = \frac{4}{u} \] Thus, \[ I = \frac{1}{4} \int \frac{\log u}{u} \, du \] 

Step 2: Integration 
Using the identity, \[ \int \frac{\log u}{u} \, du = \frac{(\log u)^2}{2} \] Thus, \[ I = \frac{1}{4} \cdot \frac{(\log(1 + x^4))^2}{2} = \frac{(\log(1 + x^4))^2}{8} \] 

Step 3: Identifying \(f(x)\), \(g(x)\), and \(h(x)\) 
From the given format, - \( f(x) = \frac{1}{8} \) - \( g(x) = 1 + x^4 \) - \( h(x) = x^2 \) 

Step 4: Computing \(h(x) [f(x) + f(\frac{1}{x})] \) 
Since \( f(x) = \frac{1}{8} \), and \( f\left(\frac{1}{x}\right) = \frac{1}{8} \), \[ h(x) [f(x) + f\left(\frac{1}{x}\right)] = x^2 \left( \frac{1}{8} + \frac{1}{8} \right) \] \[ = x^2 \cdot \frac{2}{8} = \frac{x^2}{4} \] Since \( g(x) = 1 + x^4 \), this simplifies to: \[ \frac{g(x)}{2} \] 

Final Answer: (B) \( \frac{g(x)}{2} \)