Question

\(∫^1_0\dfrac{x}{x^2-4}dx=?\)

• A. \(\dfrac{-2\pi}{6}\)
• B. \(\ln \frac{\sqrt{3}}{2}\)
• C. \(ln(\dfrac{3}{2})\)
• D. \(ln(\dfrac{√3}{2})\)
• E. \(ln(\dfrac{3}{√2})\)

Answer 1

The Correct Option is B

The integral we need to evaluate is:

\[ \int_0^1 \frac{x}{x^2 - 4} dx \]

Let \( u = x^2 - 4 \). Then \( du = 2x dx \), and \( x dx = \frac{1}{2} du \).

When \( x = 0 \), \( u = 0^2 - 4 = -4 \).

When \( x = 1 \), \( u = 1^2 - 4 = -3 \).

The integral becomes:

\[ \int_{-4}^{-3} \frac{1}{2u} du = \frac{1}{2} \int_{-4}^{-3} \frac{1}{u} du = \frac{1}{2} [\ln |u|]_{-4}^{-3} \]

\[ = \frac{1}{2} (\ln|-3| - \ln|-4|) = \frac{1}{2} (\ln 3 - \ln 4) = \frac{1}{2} \ln \frac{3}{4} = \ln \sqrt{\frac{3}{4}} = \ln \frac{\sqrt{3}}{2} \]

Answer 2

Substitution:

• Let \( u = x^2 - 4 \), so that \( du = 2x \, dx \) or \( x \, dx = \frac{1}{2} du \).
• When \( x = 0 \), \( u = 0^2 - 4 = -4 \).
• When \( x = 1 \), \( u = 1^2 - 4 = -3 \).

The integral becomes:

\[ \int_{0}^{1} \frac{x}{x^2 - 4} \, dx = \int_{-4}^{-3} \frac{1}{u} \cdot \frac{1}{2} \, du = \frac{1}{2} \int_{-4}^{-3} \frac{1}{u} \, du. \]

Evaluate the integral:

• The integral of \( \frac{1}{u} \) is \( \ln|u| \).
• Substitute the limits:
• Evaluate the logarithmic terms:
• Since \( |-3| = 3 \) and \( |-4| = 4 \), this simplifies to:

Simplify using logarithmic properties:

• Using the property \( \ln a - \ln b = \ln\left(\frac{a}{b}\right) \):
• This can also be written as:

The correct option is (B):  \( \ln\left(\frac{\sqrt{3}}{2}\right) \).