<s>[INST] Solving integral with Leibniz' rule 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?
 [/INST] 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
$$\int_0^1 \frac{x}{(1 + 50 x)^2} \, dx = \frac{1}{2500} \ln (51) - \frac{1}{2550}.$$
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