Having trouble solving $\int\frac{5x^2+3x+2}{x(x+1)^2}$

Question

I've first transformed the integral to

$$\int\frac{5x^2+3x+2}{x(x^2+2x+1)}dx$$

Which gave me

$$\frac{A}{x}+\frac{Bx+C}{x^2+2x+1}$$ $$=\frac{A(x^2+2x+1)+Bx^2+Cx}{x(x^2+2x+1)}$$ $$\frac{5x^2+3x+2}{x(x^2+2x+1)}=\frac{(A+B)x^2+(2A+C)x+A)}{x(x^2+2x+1)}$$

So I've found the corresponding variables

$$A = 2$$ $$A+B = 5, B = 3$$ $$2A+C=3, C=-1$$

So the final integral is

$$2\int\frac{dx}{x}+3\int\frac{xdx}{x^2+2x+1}-\int\frac{dx}{x^2+2x+1}$$ $$=2ln(x) -ln(x^2+2x+1)+\frac{3}{x+1}+3ln(x+1)$$

However, the expected answer is

$$2ln(x)+3ln(x+1)+\frac{4}{x+1}$$

What is my error ?

Your partial fractions decomposition, though correct, is not optimal for the integration (please see the answers). But let's continue on your path. We need to find $\int\frac{3x-1}{(x-1)^2},dx$. Rewrite this as $\int\frac{3x+3}{(x+1)^2},dx-\int\frac{4}{(x+1)^2},dx$. The first integral can be done using the substitution $u=(x+1)^2$, and the second can probably be written down directly. — André Nicolas, Mar 15 '15 at 16:38
What is up with all of the downvotes on the answers to this question...? I've flagged this post for moderator attention. — Cameron Williams, Mar 15 '15 at 16:48

score 2 · Answer 1 · answered Mar 15 '15 at 16:30

2

Hint: Notice that $x^2 + 2x + 1 = (x+1)^2$ then

$$\frac{5x^2 + 3x + 2}{x(x+1)^2} = \frac{A}{x} + \frac{B}{(x+1)} + \frac{C}{(x+1)^2}$$

answered Mar 15 '15 at 16:30

Aaron Maroja

17,571

1

The last numerator should just be C. – John Brevik Mar 15 '15 at 16:31
1

This is not the question !, he did this but he made a mistake so the question is where is the mistake? – Elaqqad Mar 15 '15 at 16:32
Well this is an easier way to see it, and eventually avoid any mistakes. – Aaron Maroja Mar 15 '15 at 16:33
no that's just an excuse. – RE60K Mar 15 '15 at 16:33
@ADG What do you mean? – Aaron Maroja Mar 15 '15 at 16:34
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@AaronMaroja: Sorry, had same idea, was just typing. – Frieder Mar 15 '15 at 16:38
you didn't found his mistake and just told another way therefore. – RE60K Mar 15 '15 at 16:40
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@ADG And in what world this answer is not useful? – Aaron Maroja Mar 15 '15 at 16:43
I just replied on your reply to Elaqqad and never said it was useless. – RE60K Mar 15 '15 at 16:44
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I don't understand people here, seriously, too many downvotes without even knowing what useful means. That's cool though. – Aaron Maroja Mar 15 '15 at 16:45
@AaronMaroja, it is far from cool. – abel Mar 15 '15 at 17:05
@abel I know, I didn't want to seem desperate for votes though, there must be some lack of judgement, I don't know. – Aaron Maroja Mar 15 '15 at 17:47

RE60K · Accepted Answer · 2015-03-15T16:45:29.563

2

I found your mistake $$\scriptsize2\int\frac{dx}{x}+3\int\frac{xdx}{x^2+2x+1}-\int\frac{dx}{x^2+2x+1}\color{red}{\ne}2ln(x) -ln(x^2+2x+1)+\frac{3}{x+1}+3ln(x+1)$$ But: $$2\int\frac{dx}{x}=2\ln|x|+c\\ 3\int\frac{xdx}{x^2+2x+1}\stackrel{t=x+1}=3\int\frac{(t-1)dt}{t^2}=3\int\left(\frac1t-\frac1{t^2}\right)dt=3\ln|x+1|+\frac3{x+1}+c\\ -\int\frac{dx}{x^2+2x+1}=\frac1{x+1}+c$$ Add them.

edited Mar 15 '15 at 16:45

answered Mar 15 '15 at 16:33

RE60K

17,716

Exactly what I needed, thanks a lot. – student Mar 15 '15 at 17:04

score 2 · Answer 3 · answered Mar 15 '15 at 16:36

2

You may use this:

$$\frac{5 x^2+3 x+2}{x (x+1)^2}=\frac{2}{x}+\frac{3}{x+1}-\frac{4}{(x+1)^2}$$

answered Mar 15 '15 at 16:36

Frieder

1,655

score 2 · Answer 4 · answered Mar 15 '15 at 16:43

2

you can see that $$\frac{5x^2 + 3x+2}{x(x+1)^2} = \frac 2 x + \frac{B}{x+1} - \frac 4{(x+1)^2}$$ by looking at the behavior of the two sides near the singularities $x = 0, x = -1.$ to find $B,$ put $x = 1.$ you will find $B = 3.$

you can now integrate $$ \int \frac{5x^2 + 3x+2}{x(x+1)^2} \, dx = 2 \ln x + 3 \ln(x+1) + \frac 4{x+1} + C$$

answered Mar 15 '15 at 16:43

abel

29,170

Where does the 2 and 4 comes from ? – student Mar 15 '15 at 17:01
1

see what happens when $x$ is close to $0.$ the left hand side is $\frac{5 \times 0^2 + 3 \times 0 + 2}{x \times 1^2} = \frac{2}{x} + \cdots.$ you can do the same when $x$ is close to $-1$ – abel Mar 15 '15 at 17:04

Having trouble solving $\int\frac{5x^2+3x+2}{x(x+1)^2}$

4 Answers4