∫1/(2x-5)²dx=? (-1/4x-10+c) ∫secxdx=? (lns|secx+tanx|+c)
yihuuuu integral gelmiş 1. 2x-5=u 2.dx=du 1/2∫2.[1/(u)²]dx 1/2∫du/u² 1/2∫(u⁻²)du 1/2.(u⁻¹)/(-1)+c 1/2.(-1/(2x-5))+c
2) https://img254.imageshack.us/img254/...at13042013.jpg