Integrate ∫(x-4)/x^2 dx from 1 to 2

Please show how you solved it. Thank you!

Rewritten, this is

∫ (x-4) * x^-2 dx....................Multiplicative form (as opposed to fractional form)
=∫ x^-1 - 4x^-2 dx.................Distributed
=[ln(x) + 4/x] (1->2)..............Integral of 1/x is ln(x); integral of -4/x^2 is 4/x
=(ln(2) + 2) - (ln(1) + 4).........Plug in bounds

=[ln(2) - 2]...........................Final Answer

Proof using wolframalpha.com
http://www.wolframalpha.com/input/?i=%E2%88%AB%28x-4%29%2Fx%5E2+dx+from+1+to+2

that is the same as INT (x-4)(x^1/2) dx

integration by parts: u=x-4 du=dx dv=x^(1/2) v=(2/3)(x^(3/2))

uv-INTv du = (x-4)(2/3)(x^(3/2)) - INT (2/3)(x^(3/2))dx

throw the 2/3 out front: (x-4)(2/3)(x^(3/2)) - (2/3) INT (x^(3/2))dx

integrate x^(3/2)dx : (x-4)(2/3)(x^(3/2)) - (2/3) [(2/5)(x^(5/2))]

then just solve the integral from 1 to 2:

[ (2(2-4)(2^(3/2))/3 - (4(2^(5/2))/15 ] - [ (2(1-4)(1^(3/2))/3 - (4(1^(5/2))/15 ]

- which equals -

[ ((-4)(2.8284))/3 - (4(5.6568))/15 ] - [ -6/3 - 4/15 ]

- which equals -

-3.0130133333333333333333333333.....3333… more 3's....

Integrate ∫(x-4)/x^2 dx from 1 to 2
Sol: ∫(1/x) -4(1/x^2) dx from 1 to 2 (Divide each term by x^2)
[log(x) +(1/x)] take limits from 1 to2
=[{log(2) - log(1)} +{1/2 -1}
= log(2)- (1/2) ...............Ans

∫(x-4)/x^2 dx from 1 to 2
= ∫ (x-4)x^-2 dx = ∫ (x^-1 – 4x^-2) dx
…………..….2
= [ln(x) + 4/x]
…………..….1
= [(ln(2) + 4/2) – (ln(1) + 4/1)]
= [(0.693 + 2) – (0 + 4)]
= 2.693 – 4
= -1.307