HW Helper P: 1,422 Here's what I think. Let: $$g(x) := \int f(x)dx \quad \mbox{and} \quad h(x):= \int f(x)dx$$. Then, you do not have g(x) - h(x) = 0, you will have g(x) - h(x) = C, where C is some constant. So here's the same, you can say that: $$\int \frac{dx}{x} - \int \frac{dx}{x} = C$$, where C is some constant. So it's not a contradiction... Viet Dao,
 HW Helper P: 1,422 Integration by Parts Contradiction You can continue integrating by parts, something like: $$\int \frac{dx}{x} = 1 + \int \frac{dx}{x} = 1 + \left( 1 + \int \frac{dx}{x} \right) = 1 + 1 + .. + 1 + \left( 1 + \int \frac{dx}{x} \right)$$. So you'll have: $$\Leftrightarrow \int \frac{dx}{x} - \int \frac{dx}{x} = 1 + 1 + 1 + ... + 1$$. Viet Dao,