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

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- Feb 3rd 2007, 06:42 PMclockinglyintegration 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 - Feb 3rd 2007, 07:09 PMAfterShock
- Feb 3rd 2007, 07:10 PMThePerfectHacker
You have,

$\displaystyle \int \sqrt{x}\ln x dx$

Let,

$\displaystyle u=\ln x$ and $\displaystyle v'=\sqrt{x}=x^{1/2}$

Then,

$\displaystyle u'=1/x$ and $\displaystyle v=(2/3)x^{3/2}$

Thus,

$\displaystyle \frac{2}{3}\ln x\cdot x^{3/2}-\frac{2}{3}\int \frac{x^{3/2}}{x} dx$

Simplify,

$\displaystyle \frac{2}{3}\ln x \cdot x^{3/2}-\frac{2}{3} \int x^{1/2} dx$

You can do it from there. - Feb 3rd 2007, 07:13 PMticbol
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)] -