Posted by Anon on Monday, December 3, 2012 at 11:43am.
x (2x-1)^.5 dx
u = x
dv = (2x-1)^.5 dx
du = dx
v = (1/3)(2x-1)^1.5
so
(x/3)(2x-1)^(1.5) -(1/3) (2x-1)^1.5 dx
(x/3)(2x-1)^1.5-(1/3)(1/5)(2x-1)^2.5
(1/3)[ x(2x-1)^1.5 - (1/5)(2x-1)^2.5 ]
Related Questions
Calculus - Evaluate integral of x times the square root of (x-4) dx I got u-(8u...
Calculus - Find the exact area of the region enclosed by the square root of (x...
Calculus - (integral) e^3x dx A. e^3x+C B. 1/3e^3x+C C. e^4x+C D. 1/4e^4x+C ...
calculus - 6. What is the derivative of y = square root of 3x ? a.dy/dx = square...
Calculus - evaluate the integral of dx/square root of 9-8x-x^2
Calculus - Evaluate the integral from 0 to 24 of the square root of 9+3x dx.
Calculus - Evaluate the integral from [(2/square rt 2),0] of dx/sqaure root of (...
Business Calculus - Evaluate each indefinite integral. Use the substitution ...
Maths Calculus Derivative Integral - Urgent Please - Use 2nd Fundamental Theorem...
Math/Calculus - How would I evaluate the following integral by using integration...
For Further Reading
