∫1/(2x-5)²dx=? (-1/4x-10+c)
∫secxdx=? (lns|secx+tanx|+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
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
Süleyman Oymak 04:17 13 Nis 2013 #3 2)
