#F'(x) = 2 sqrt((2x)^2 + 2x)#
#= 2 sqrt(4x^2 + 2x)#
#=> F'(1) = 2 sqrt(6)#
#"So we are looking for the straight line with slope "2 sqrt(6)#
#"that passes through (1, F(1))."#
#"The problem is that we do not know F(1) unless we calculate"#
#"the definite integral"#
#int_1^2 sqrt(t^2 + t)" "dt#
#"We have to apply a special substitution to solve this integral."#
#"We can get there with the substitution "u - t = sqrt(t^2 + t)#
#=> (u - t)^2 = t^2 + t => u^2 - 2 u t + cancel(t^2) = cancel(t^2) + t#
#=> t = u^2/(1+2u)#
#=> dt/{du} = (2u(u+1))/(1+2u)^2#
#=> t^2 + t = u^4/(1+2u)^2 + u^2/(1+2u) = ((u(u+1))/(1+2u))^2#
#=> sqrt(t^2+t) = (u(u+1))/(1+2u)#
#t = 1 => u^2 - 2u - 1 = 0 => u = 1 + sqrt(2)#
#t = 2 => u^2 - 4 u - 2 = 0 => u = 2 + sqrt(6)#
#"(we take the solution with + sign because "u - t = sqrt(...) > 0")"#
#int sqrt(t^2 + t) " "dt = 2 int u^2(u+1)^2/(1+2u)^3" "du#
#= 2 int (u^4 + 2 u^3 + u^2) / (8 u^3 + 12 u^2 + 6 u + 1)" "du#
#= 2 int (u/8 + 1/16)" "du - 2 int (u^2/2 + u/2 + 1/16)/(1+2u)^3" "du#
#= 2 (u^2 + u)/16 - 2 int (A/(1+2u) + B/(1+2u)^2 + C/(1+2u)^3)" "du#
#" (splitting up in partial fractions)"#
#=> A = 1/8, B = 0, C = -1/16#
#= 2 (u^2 + u)/16 - 2 ln(|1+2u|)/16 - 2 /(64(1+2u)^2)#
#= (u^2 + u) / 8 - ln(|1 + 2u|) / 8 - 1/(32(1+2u)^2)#
#"We evaluate between "u = 1+sqrt(2)" and "u = 2+sqrt(6)#
#"and we obtain the value"#
#F(1) = 1.935#
