Thread: integration by parts

hi, i have to integrate the following problem by parts. i started to do it, but then i got stuck halfway through. if anyone could show me how it is done, i would be grateful.

INT (square root of y) * (ln(y)) dy

Originally Posted by clockingly

hi, i have to integrate the following problem by parts. i started to do it, but then i got stuck halfway through. if anyone could show me how it is done, i would be grateful.

INT (square root of y) * (ln(y)) dy

Let u = ln(y)
du = 1/y

dv = sqrt(y)
v = (2*y^(2/3))/3

uv - int(vdu)

[2*y^(3/2)*ln(y)]/3 - int((2*sqrt(y))/3)

[2*y^(3/2)*ln(y)]/3 - (4*y^(3/2))/9 + C

Which is equivalent to:

[2*y^(3/2)*(3*ln(y) - 2)]/9 + C

Originally Posted by clockingly

hi, i have to integrate the following problem by parts. i started to do it, but then i got stuck halfway through. if anyone could show me how it is done, i would be grateful.

INT (square root of y) * (ln(y)) dy

You have,

Let,
and
Then,
and
Thus,

Simplify,

You can do it from there.

Originally Posted by clockingly

hi, i have to integrate the following problem by parts. i started to do it, but then i got stuck halfway through. if anyone could show me how it is done, i would be grateful.

INT (square root of y) * (ln(y)) dy

By parts.
INT.[u]dv = u*v -INT.[v]du -------(i)

INT.[sqrt(y) *ln(y)]dy -------------(1)

You can try two ways. I did, and I found this is the correct assumption.
Let u = ln(y)
Hence, du = (1/y)dy

And so, dv = sqrt(y) dy = (y^(1/2))dy
Hence, v = 1/(3/2) *y^(3/2) = (2/3)y^(3/2)

INT.[sqrt(y) *ln(y)]dy -------------(1)
Apply the assumptions, and (i),
= ln(y)*[(2/3)y^(3/2)] -INT.[(2/3)y^(3/2)][(1/y)dy]
= (2/3)(y^(3/2))ln(y) -(2/3)INT.[y^(1/2)]dy
= (2/3){y^(3/2) ln(y) -(2/3)(y^(3/2))} +C
= [(2/3)y^(3/2)]{ln(y) -2/3} +C
= (2/3)(y)(sqrt(y))[ln(y) -2/3] +C --------------answer.
= [ln(y) *(2/3)y^(3/2)] -