integral_0^0.4 1/(0.02+0.025 x)^2 dx = 666.667 how to solve this manually?? help
int [0 .. 0.4] 1/(0.02+0.025 x)^2 dx.
Consider the integral of 1/(a+bx)^n.
This is simply (-1/(b(n-1)))*(1/(a+bx)^(n-1))
(It is easier to inderstand the above if you differentiate it to get the original expression.)
if a = 2, b = 3 and n = 4, then int 1/(2+3x)^4 = (-1/(3*3))*(1/(2+3x)^(3))
So, when we have 1/(0.02+0.025 x)^2
the integral is: (-1/(0.025*1)*(1/(0.02 + 0.025x)^(1))
this simplifies as: -40/(0.02 + 0.025x)
Now simply enter your limits to get the result 666.667