Question

\(∫^1_0\dfrac{x}{x^2-4}dx=?\)

• A. \(\dfrac{-2\pi}{6}\)
• B. \(\dfrac{1}{2}ln(\dfrac{4}{3})\)
• C. \(ln(\dfrac{3}{2})\)
• D. \(ln(\dfrac{√3}{2})\)
• E. \(ln(\dfrac{3}{√2})\)

Answer 1

The Correct Option is B

The correct option is (A): \(\dfrac{1}{2}ln(\dfrac{4}{3})\).
\(∫^1_0\dfrac{x}{x^2-4}dx\)
To solve the above expression
first take,  \(x^{2}-4= t\)
Now derivate both the sides w.r.t \(x\)
we get,
\(2x dx= dt\)
\(xdx=\dfrac{1}{2}dt\)
By substituting the term in the parent expression we can write
\(\dfrac{1}{2}∫^1_0\dfrac{dt}{t}\)
\(=\dfrac{1}{2}ln(t)|_0^1\)
\(=\dfrac{1}{2}ln((x^{2})-4)|_0^1\)
by applying the limits we get,
\(=\dfrac{1}{2}[ln(1-4)-ln(0-4)]\)
\(=\dfrac{1}{2}ln(\dfrac{4}{3})\)