Question

Evaluate : \( \int_{0}^{1} \log(1 + x^2)dx \)

Hint:

Always remember to evaluate limits carefully for the term \( uv \) in integration by parts.
When the numerator and denominator of a rational function have the same degree, divide or use the \( +1, -1 \) trick.

Answer 1

Concept:
• Integration by Parts: \( \int u \cdot dv = uv - \int v \cdot du \).
• For logarithmic functions, we take the log term as \( u \) and \( 1 \) as \( dv \).
• Substitution method and algebraic adjustment for rational integrals.

Step 1: Apply Integration by Parts
Let \( I = \int_{0}^{1} \log(1 + x^2) \cdot 1 \, dx \). Taking \( u = \log(1 + x^2) \) and \( dv = dx \), we have \( du = \frac{2x}{1 + x^2}dx \) and \( v = x \). \[ I = \left[ x \log(1 + x^2) \right]_{0}^{1} - \int_{0}^{1} x \cdot \frac{2x}{1 + x^2} \, dx \]

Step 2: Evaluate the boundary terms and simplify the remaining integral
Evaluating the first part: \[ \left[ 1 \cdot \log(1 + 1^2) \right] - \left[ 0 \cdot \log(1 + 0^2) \right] = \log 2 - 0 = \log 2 \] Now the integral part becomes: \[ -2 \int_{0}^{1} \frac{x^2}{1 + x^2} \, dx \]

Step 3: Solve the integral of the rational function
Use algebraic manipulation: \( \frac{x^2}{1 + x^2} = \frac{x^2 + 1 - 1}{1 + x^2} = 1 - \frac{1}{1 + x^2} \). \[ -2 \int_{0}^{1} \left( 1 - \frac{1}{1 + x^2} \right) \, dx = -2 \left[ x - \tan^{-1}x \right]_{0}^{1} \] \[ = -2 \left[ (1 - \tan^{-1}1) - (0 - \tan^{-1}0) \right] \] \[ = -2 \left[ 1 - \frac{\pi}{4} \right] = -2 + \frac{\pi}{2} \]

Step 4: Combine the results for the final answer
\[ I = \log 2 - 2 + \frac{\pi}{2} \]