Integrate 6/(e^x+2) between 1 and 0

By Trapezium Rule (with 5 segments)x 0 0.2 0.4 0.6 0.8 1y 2 1.8625 1.71830 1.56981 1.41994 1.27165≈1⁄2*(0.2) [2+1.27165+ 2( 1.8625 + 1.71830 + 1.56981 + 1.41994)]≈1⁄10 *16.41283 = 1.641By Substitution/ Partial Fractions and Calculuslet u=e^xdu⁄dx = e^x therefore dx= 1/e^x duHence∫^1_0(6/(e^x+2))dx = ∫^b_a (6/u(u+2))dua=e^0=1 and b=e^1=eTherefore∫^1_0(6/(e^x+2))dx = ∫^e_1 (6/u(u+2))du6/u(u+2) = A/u + B/(u+2)6= A(u+2) +B(u)Let u=0, A=3Let u=-2 B=-3Therefore∫^e_1 (6/u(u+2))du = ∫^e_1 (3/u+ 3/(u+2))du= [3ln(u) - 3ln(u+2)]^e_1= [3ln(e) - 3ln(e+2)]-[3ln1 - 3ln3]=3 - 3ln(e+2) +3ln3= 3 -3ln(3/(e+2))