Use integration by parts to find the value of definite integral between 5 and 1 (3x/root(2x-1))dx

use the formula integral(uv') = uv - integral(vu') so the tricky integral can be avoided (this formula can be derived from integrating the product rule) so in these questions all you need to look for is what can be differentiated out (might take more than one differentiation though).If the question is put in the form:integral(3x(2x-1)^-0.5)dxthe term we can differentiate out an x from becomes clearer, its the 3x which will be the u in the equation and therefore the (2x-1)^-0.5 will be the vlooking at the equation (2x - 1)^-0.5 is now v'. it needs to be integrated using the reverse chain rule in order to find v:v' = (ax+b)ⁿv = (ax+b)ⁿ⁺¹/a(n+1)
v' = (2x - 1)^-0.5 u = 3x v = 20.5(2x - 1)^0.5 u' = 3
now plug the values into the equation:[3x(2x-1)^0.5 -integral(3(2x-1)^0.5)]⁵₁which now can be easily integrated using the reverse chain rule again:[3x(2x-1)^0.5 -(3/3(2x-1)^1.5)]⁵₁then just sub in x=5 and x =1 and subtract for the definite integral:(15*3 - 27) - (3-1)=16