Question

Evaluate: \( \int_0^1 \frac{2x}{5x^2 + 1} dx \)

• A. \( \frac{1}{5} \log 6 \)
• B. \( \frac{1}{5} \log 5 \)
• C. \( \frac{1}{2} \log 6 \)
• D. \( \frac{1}{2} \log 5 \)

Hint:

If the derivative of the denominator is present in the numerator (up to a constant), substitution leads to a logarithmic form.

Answer 1

The Correct Option is A

Let’s use substitution. Let: \[ u = 5x^2 + 1 \Rightarrow \frac{du}{dx} = 10x \Rightarrow dx = \frac{du}{10x} \] But easier is to notice: \[ \int \frac{2x}{5x^2 + 1} dx \] Use substitution: \( u = 5x^2 + 1 \Rightarrow du = 10x dx \Rightarrow \frac{du}{5} = 2x dx \)
So the integral becomes: \[ \int \frac{2x}{5x^2 + 1} dx = \frac{1}{5} \int \frac{1}{u} du = \frac{1}{5} \log|u| + C = \frac{1}{5} \log(5x^2 + 1) + C \] Now evaluate from 0 to 1: \[ = \frac{1}{5} [\log(5(1)^2 + 1) - \log(5(0)^2 + 1)] = \frac{1}{5} [\log(6) - \log(1)] = \frac{1}{5} \log 6 \]