Solving integral with Leibniz' rule

Question

I'm trying to prove the following integral with Leibniz' rule: $\int_{0}^{1}{\frac{x}{(1+50x)^2}dx}$. Firstly I wrote: $\int_{0}^{1}{\frac{x}{(1+bx)^2}dx}$ then I need to find a function $f$ that $\frac{\partial f}{\partial b}(x,b)= {\frac{x}{(1+bx)^2}}$ but now what?

Answer 1

If you insist on using Leibniz' rule for differentiating under the integral sign it can be done as follows.

Let $$I(b) = \int_0^1 \frac{dx}{1 + bx}, \quad b > 0.$$ Note that $$I(b) = \frac{1}{b} \ln (1 + bx) \Big{|}_0^1 = \frac{1}{b} \ln (1 + b).$$

Also, on applying Leibniz' rule to $I(b)$ we have $$I'(b) = -\int_0^1 \frac{x}{(1 + bx)^2} \, dx,$$ and we see that the required integral comes from setting $b = 50$ in the above integral. Now $$I'(b) = \frac{d}{db} \left (\frac{\ln (1 + b)}{b} \right ) = \frac{1}{b(1 + b)} - \frac{1}{b^2} \ln (1 + b),$$ yielding $$\int_0^1 \frac{x}{(1 + bx)^2} \, dx = \frac{1}{b^2} \ln (1 + b) - \frac{1}{b(1 + b)}.$$

Finally, setting $b = 50$ gives $$\int_0^1 \frac{x}{(1 + 50 x)^2} \, dx = \frac{1}{2500} \ln (51) - \frac{1}{2550}.$$