Question

Evaluate: \[ \int \frac{x^2}{1 + (x^3)^2} \, dx \] 

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

Hint:

When you see a power of $x$ in the numerator that is exactly one less than the power inside a bracket in the denominator, substitution is almost always the intended path.

Answer 1

The Correct Option is C

Concept: This integration problem is best solved using the substitution method (u-substitution). We look for a part of the integrand whose derivative is also present in the expression. The standard integral formula we aim to use is: \[ \int \frac{1}{1 + u^{2}} \, du = \tan^{-1}(u) + c \]

Step 1: Perform u-substitution.
Let \( u = x^{3} \). Differentiating both sides with respect to \( x \): \[ \frac{du}{dx} = 3x^{2} \implies du = 3x^{2} \, dx \implies \frac{1}{3} du = x^{2} \, dx \]

Step 2: Substitute and integrate.
Substitute \( x^{3} = u \) and \( x^{2} \, dx = \frac{1}{3} du \) into the original integral: \[ \int \frac{x^{2}}{1 + (x^{3})^{2}} \, dx = \int \frac{1}{1 + u^{2}} \cdot \frac{1}{3} \, du \] \[ = \frac{1}{3} \int \frac{1}{1 + u^{2}} \, du \] \[ = \frac{1}{3} \tan^{-1}(u) + c \]

Step 3: Back-substitute to get the final answer.
Replace \( u \) with \( x^{3} \): \[ = \frac{1}{3} \tan^{-1}(x^{3}) + c \]