Question

Evaluate the integral \[ \int \frac{x^5}{x^2 + 1} dx. \]

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

Hint:

For rational integrals, split terms and use substitution if needed.

Answer 1

The Correct Option is B

Step 1: Splitting the integral Rewriting: \[ I = \int \frac{x^5}{x^2 + 1} dx. \] Splitting: \[ I = \int x^3 \cdot \frac{x^2}{x^2 + 1} dx. \] Step 2: Using substitution Let \( u = x^2 + 1 \), then \( du = 2x dx \). This simplifies the integral, leading to: \[ I = \frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{2} \log(x^2 + 1) + c. \]

Answer 2

Evaluate the integral:
\[ I = \int \frac{x^5}{x^2 + 1} \, dx \]

Step 1: Perform polynomial division since the degree of numerator is higher than denominator:
Divide \( x^5 \) by \( x^2 + 1 \):
\[ \frac{x^5}{x^2 + 1} = x^3 - x + \frac{x}{x^2 + 1} \]

Step 2: Rewrite the integral:
\[ I = \int \left( x^3 - x + \frac{x}{x^2 + 1} \right) dx = \int x^3 \, dx - \int x \, dx + \int \frac{x}{x^2 + 1} \, dx \]

Step 3: Integrate each term separately:
\[ \int x^3 \, dx = \frac{x^4}{4} \] \[ \int x \, dx = \frac{x^2}{2} \] \[ \int \frac{x}{x^2 + 1} \, dx \] For the last integral, let \( u = x^2 + 1 \), then \( du = 2x dx \Rightarrow x dx = \frac{du}{2} \), so:
\[ \int \frac{x}{x^2 + 1} dx = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \log|u| + C = \frac{1}{2} \log(x^2 + 1) + C \]

Step 4: Combine results:
\[ I = \frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{2} \log(x^2 + 1) + C \]

Therefore, the integral evaluates to:
\[ \boxed{\frac{x^4}{4} - \frac{x^2}{2} + \frac{1}{2} \log(x^2 + 1) + C} \]