Originally Posted by

**simplependulum** If my calculations are correct , the integrals are both zero .

For the first one ,

Sub. $\displaystyle x = \frac{ab}{t} $ in the second integral .

and we will find that it is equal to

$\displaystyle \int_a^b \frac{e^{x/a}}{x}~dx - \int_a^b \frac{e^{t/a}}{t}~dt $

And the second integral , i find that it is identical to

$\displaystyle \sqrt[3]{(2x-1)[(x-\frac{1}{2})^2 - \frac{5}{4}]} $

Sub. $\displaystyle x = 1-t $ and we have

$\displaystyle I = -I $

$\displaystyle I = 0 $