∫dx(2x−7)√(x−3)(x−4)
∫dx(2x−7)√x2−7x+12
u=√x2−7x+12
dx=2√x2−7x+12(2x−7)dx
2∫du4u2+1
2tan−1(2u)
2tan−1(2√x2−7x+12)+C