Question

Evaluate the integral: $$ \int \frac{x^3 \tan^{-1}(x^4)}{1 + x^8} \,dx. $$ 

• A. \( \frac{(\tan^{-1}(x^4))^2}{8} + c \)
• B. \( \frac{(\tan^{-1}(x^4))^3}{3} + c \)
• C. \( \frac{(\tan^{-1}(x^4))^2}{4} + c \)
• D. \( \frac{(\tan^{-1}(x^4))^2}{2} + c \)

Hint:

For integration involving inverse trigonometric functions, check if differentiation of the inverse function appears in the numerator.

Answer 1

The Correct Option is A

Step 1: Substituting \( u = \tan^{-1}(x^4) \) Differentiate both sides: \[ du = \frac{4x^3}{1 + x^8} dx. \] Rearranging: \[ x^3 dx = \frac{(1 + x^8) du}{4}. \] Substituting in the given integral: \[ \int u \cdot \frac{(1 + x^8) du}{4}. \] \[ \frac{1}{4} \int u \, du. \]
Step 2: Solving the Integral \[ \frac{1}{4} \cdot \frac{u^2}{2} = \frac{u^2}{8}. \] \[ = \frac{(\tan^{-1}(x^4))^2}{8} + c. \] Thus, the correct answer is option (1).

Answer 2

To evaluate the integral \( \int \frac{x^3 \tan^{-1}(x^4)}{1 + x^8} \,dx, \) we will use integration by parts.

Let's denote:

• \( u = \tan^{-1}(x^4), \) then the derivative \( du \) is given by:
  \( du = \frac{d}{dx}[\tan^{-1}(x^4)] \, dx = \frac{4x^3}{1+(x^4)^2} \, dx = \frac{4x^3}{1+x^8} \, dx. \)
• \( dv = \frac{x^3}{1+x^8} \, dx, \) hence integrate to find:
  \( v = \int \frac{x^3}{1+x^8} \, dx. \)
  To integrate \( v \), set \( w = x^4 \), \( dw = 4x^3 \, dx \), therefore \( dx = \frac{dw}{4x^3} \).

Substitute into the integral:

\( v = \frac{1}{4} \int \frac{dw}{1+w^2} = \frac{1}{4} \tan^{-1}(w) + C = \frac{1}{4} \tan^{-1}(x^4) + C. \)

Now apply integration by parts:

\( \int u \, dv = uv - \int v \, du \)

• \( uv = \tan^{-1}(x^4) \cdot \frac{1}{4} \tan^{-1}(x^4) = \frac{1}{4} (\tan^{-1}(x^4))^2. \)
• \( \int v \, du = \int \left(\frac{1}{4} \tan^{-1}(x^4)\right) \left(\frac{4x^3}{1+x^8}\right) dx = \int \frac{x^3 \tan^{-1}(x^4)}{1+x^8} dx. \)

Since \( \int v \, du = \int \frac{x^3 \tan^{-1}(x^4)}{1+x^8} dx, \) this integral was our original integral. Thus the integral evaluates to:

\( I = \frac{1}{2} \times \frac{1}{4} (\tan^{-1}(x^4))^2 + C = \frac{1}{8} (\tan^{-1}(x^4))^2 + C. \)

Thus, the result is:

\( \boxed{\frac{(\tan^{-1}(x^4))^2}{8} + c} \)