Evaluate $\int_2^\infty{\frac{3x-2}{x^2(x-1)}}$

Question

To be shown that $\int_2^\infty{\dfrac{3x-2}{x^2(x-1)}}=1-\ln2$

My thought: $\dfrac{3x-2}{x^2(x-1)}=\dfrac{3x}{x^2(x-1)}-\dfrac{2}{x^2(x-1)}$

• $\dfrac{3x}{x^2(x-1)}=\dfrac{3}{x(x-1)}=\ldots=-\dfrac{3}{x}+\dfrac{3}{x-1}$

• $\dfrac{2}{x^2(x-1)}=\ldots=-\dfrac{2}{x^2}+\dfrac{2}{x-1}$

But developing the sum of the integrals of the above two gives ln of infinite in my results. I don't know what I am doing wrong. Can you please confirm the dots above? Although I have several times. Thanks a lot

There is some not right algebra, entry 2. Then there is a little trick needed once the algebra is fixed. — André Nicolas, Jul 13 '13 at 18:47
@darkchampionz: Yes, but can you say that of item 2? Calculate from your $\frac{-2}{x^2}+\frac{2}{x-1}$ backwards. — André Nicolas, Jul 13 '13 at 18:58

score 7 · Accepted Answer · answered Jul 13 '13 at 18:45

7

Hint

Write the integrand as: $$\frac{3x-2}{x^2(x-1)}=\frac{Ax+B}{x^2}+\frac{C}{x-1}$$ in which $A,B$ and $C$ is unknown constants and so find the proper values for them. I think via this way you can find them easier than the way you noted. Indeed, $$\frac{Ax+B}{x^2}+\frac{C}{x-1}=\frac{(A+C)x^2+x(B-A)-B}{x^2(x-1)}$$ and therefore you get $$A+C=0, ~~B-A=3,~~-B=-2$$ and so...

answered Jul 13 '13 at 18:45

Mikasa

67,374

2

Thanks a lot but what about my way? Because I can't find something wrong in it :| – darkchampionz Jul 13 '13 at 18:54
Babak S.: $\color{red}{+1}$ – M.H Jul 13 '13 at 18:56
Doesn't get any simpler...+1 – DonAntonio Jul 13 '13 at 18:56
@DonAntonio: Thanks. If you don't mind, I use the form "..." as you have been doing. :-) – Mikasa Jul 13 '13 at 19:03
@darkchampionz: Dear friend, following Andre's comments will lead you right. – Mikasa Jul 13 '13 at 19:13
1

Nice work, Babak! ;-) – amWhy Jul 14 '13 at 11:55

score 0 · Answer 2 · answered Jul 14 '13 at 15:51

HINT:

Using Partial Fraction Decomposition,

$$\dfrac{3x-2}{x^2(x-1)}=\frac A{x^2}+\frac Bx+\frac C{x-1}$$

$$\implies 3x-2=A(x-1)+Bx(x-1)+Cx^2=(B+C)x^2+x(A-B)-A$$

Comparing the constants in the above identity $A=2$

Comparing the coefficients of $x,A-B=3\implies B=A-3=-1$

Comparing the coefficients of $x^2,B+C=0\implies C=-B=1$

Alternatively, $$ 3x-2=A(x-1)+Bx(x-1)+Cx^2$$

Putting $x=1$ in the above identity, $C=3\cdot 1-2=1$

Putting $x=0,-2=A(0-1)\implies A=2$

Comparing the coefficients of $x^2,B+C=0\implies B=-C=-1$

Evaluate $\int_2^\infty{\frac{3x-2}{x^2(x-1)}}$

2 Answers2