1) INT.(2->1)[lnX / X^2]dX --------------(1)Originally Posted bynirva

By parts.

INT.[u]dv = uv -INT.[v]du ------(i)

Let u = lnX

so, du = (1/X)dX

And dv = dX / X^2

So, v = -1/X

Then, for the indefinite integration of (1),

INT.[lnX / X^2]dX

= (lnX)(-1/X) -INT.[-1/X](dX /X)

= -(lnX)/X +INT.[1/X^2]dx

= -(lnX)/X +(-1/X) +C

= -(LnX)/X -1/X +C

And so for the definite integration of (1),

= -[(lnX)/X +1/X](2->1)

= -[ln(1)/1 +1/1] +[ln(2)/2 +1/2]

= -1 +(1/2)ln(2) +1/2

= (1/2)[ln(2) - 1] -----------answer.

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2) INT.[cos(lnX)]dx -----(2)

Whoa, this could be long, so let me skip some steps.

Let U = lnX

So, X = e^U

and, dU = dX /X

dX = X*dU = e^U dU

Then, substitute those into (2),

INT.[cos(lnX)]dX

= INT.[cosU](e^U dU) ------(2a)

Integrate (2a) by parts:

INT.[r]ds = r*s -INT.[s]dr

r = e^U; so, dr = e^U dU

ds = cosU dU; so, s = sinU

Then, back to (2a),

= e^U *sinU -INT.[sinU](e^U dU)

= e^U *sinU -INT.[e^U](sinU dU) -------(2b)

Integrate the last part of (2b) by parts again,

INT.[m]dn = m*n -INT.[n]dm

m = e^U; so, dm = e^U dU

dn = sinU dU; so, n = -cosU

Then, back to (2b),

= e^U *sinU -{e^U *(-cosU) -INT.[-cosU](e^U dU)}

= e^U *sinU +e^U *cosU -INT.[cosU](e^U dU) -------(2c)

The last part of (2c) is the same as (2a),

And since (2a) is (2) also, then,

= e^U *sinU +e^U *cosU -INT.[cos(lnX)]dX

Transpose the last part to the unseen LHS,

2(INT.[cos(lnX)]dX) = e^U *sinU +e^U *cosU

So, divide both sides by 2,

INT.[cos(lnX)]dX = (1/2)[e^U *sinU +e^U *cosU]

Substituting back away from U,

INT.[cos(lnX)]dX = (1/2)[X*sin(lnX) +X*cos(lnX)]

Therefore,

INT.[cos(lnX)]dX = (X/2)[sin(lnX) +cos(lnX)] +C. --------answer

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For general derivatives or integrals, go to this website:

http://www.sosmath.com/tables/tables.html

(Sorry, I do not know how to link.)

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Hey, it automatically linked. Amazing.