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

Answer: (B)

$\dfrac{1}{2}ln(\dfrac{4}{3})$

Answer 2

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,
$=\dfrac{1}{2}[ln(1-4)-ln(0-4)]$
$=\dfrac{1}{2}ln(\dfrac{4}{3})$