Please solve this for me..

∫ln(2x) dx   ...we can do this one by parts....(it's pretty simple, really!!!!)

let dv = (1) dx    v = x

let u = ln(2x)   du = 2/(2x) dx  = 1/x dx

So we have

∫ u dv =  [u * v ] - ∫ v du  =  x*ln(2x) - ∫ x (1/x) dx  =  x*ln(2x)  - ∫  dx =  x*ln(2x) - x

And  substituting in the integration limits we have

[ 2ln[2(2)] - 1 ln[2(1)] - [2 - 1] = about 1.0794

And that's it !!!

You have got to be joking - Incidently, it is not funny!

if you open the image at new window.. The image will become upsise down back.. Maybe this person upload it through a phone.. This could be sometimes.. Dont be harsh..

Here's the second one.....the partial fraction decomposition deals with repeated linear factors, so we have

[6x - 3] /  [(x +1)^2 (x -1)] =  A / (x + 1) + B / (x + 1)^2  + C / (x - 1)

And we have

(6x - 3) = A(x+1)(x-1) + B ( x -1) + C (x+1)^2    expand

(6x - 3)  = Ax^2 - A + Bx - B + Cx^2 + 2C(x) + C   simplify

(6x - 3) = (A + C)x^2 + (B + 2C)x - (A + B - C)     ....and, equating coefficients, we have the folllowing system of equations

A + C = 0  →  A =- C

B + 2C = 6

A + B - C = 3   →  -2C + B = 3

And, adding the second and third equations, we have, 2B = 9 → B = 9/2

Then, using the second equation, (9/2) + 2C = 6 →  2C = 3/2 → C = 3/4 → A -3/4

So this gives us

∫ [6x-3]/ [(x+1)^2 (x-1)] dx =

∫(-3/4)/(x+1) dx + ∫(9/2)/(x + 1)^2 + ∫(3/4)/(x-1) dx

The first and last integrals become

-(3/4)ln(x + 1)  + (3/4)ln(x-1)  =  (3/4)[ln (x - 1) - ln(x+1)  = (3/4)ln [(x-1) / (x+1)] + C₁

For the middle integral, let x + 1 = u   ...then du = dx  and we have

(9/2)∫u^(-2) du = -(9/2)u^(-1)  = (-9/2)/(x + 1) + C₂

And putting  everything together (and letting C1 + C2  = C), we arrive at

(3/4)ln [(x-1) / (x+1)] - (-9/2)/(x + 1) + C

Using partial fractions can be helpful in these situations and usually leads to  "ln" integrals which can be more easily evaluated...!!!