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

3 Answers
t0hierry
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

Jim H
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.

t0hierry
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

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