How do you integrate #int xsqrt(x-1)# by parts?

3 Answers
Jan 30, 2017

#x = cos^2 y#
#sqrt(x-1) = siny#
#I = int cos^2 y siny 2cosy siny (-1)#
#I = int (-2) cos^3ysiny#
#I = 1/2 cos^4 y#

Jan 30, 2017

Please see below.

Explanation:

#intxsqrt(x-1) dx#

Let #u = x# and #dv = sqrt(x-1) dx#

so that #du=1dx# and #v = 2/3(x-1)^(3/2)#

#uv-intvdu = 2/3x(x-1)^(3/2) - 2/3 int (x-1)^(3/2) dx#

# = 2/3x(x-1)^(3/2) - 2/3 [2/5 (x-1)^(5/2)] +C#

# = 2/3x(x-1)^(3/2) - 4/15 (x-1)^(5/2) +C#.

Rewrite algebraically to taste. I like the answer above, but others might prefer

# = 2/15[5x(x-1)^(3/2) - 2 (x-1)^(5/2)]+C#

Or

# = 2/15(x-1)^(3/2)(3x+2)+C#

Or some equivalent expression.

Jan 30, 2017

Without integration by parts one can see that
#x = x - 1 + 1#
so
#x sqrt(x-1) = (x-1)sqrt(x-1) + sqrt(x-1)#
#int x sqrt(x-1) dx = int (x-1)^(3/2)d(x-1) + int (x-1)^(1/2)d(x-1) #
#int x sqrt(x-1) dx = (x-1)^(5/2)2/5 + (x-1)^(3/2)2/3#