Why I'm Getting Wrong Answer With This Method: f(x)=\int\frac{x+1}{x^2+2x}

[e^(i Pi)+1=0]

Homework Statement

$f(x)=\int\frac{x+1}{x^2+2x}$

The Attempt at a Solution

I know that this can be solved with substitution. I'm wondering why I get the wrong answer when using this other method...

$f(x)=\int\frac{x+1}{x^2+2x}$

$=\int\frac{x}{x^2+2x}+\frac{1}{x^2+2x}$


$=\int\frac{1}{x+2}+\int\frac{1}{x^2+2}$

$=ln(x+2)+ln(x^2+2) + c$

I could simplify more, but it's wrong. It should be $.5(x^2+2) + c$

I'm guessing that $\int\frac{1}{x^2+2}\neq ln(x^2+2)$ so in what cases does $\int \frac{1}{x} = ln(x) + c$ not hold?? I just learned this stuff today so any answers will have to be rather explicit please.

 

[Harrisonized]

I'm guessing that ∫ 1/(x²+2) dx ≠ ln(x²+2)

This.

d/dx ln(x²+2) = 2x/(x²+2), which is obviously not equal to 1/(x²+2). Never forget the chain rule.

To integrate 1/(x²+2), you need to use trigonometric substitution.

 

[e^(i Pi)+1=0]

Don't know that yet. So I guess that any time there's a nested function then ∫(1/x) dx=ln(x)+c doesn't hold?

 

[Harrisonized]

Do you know of the chain rule?

 

[SammyS]

Don't know that yet. So I guess that any time there's a nested function then ∫(1/x) dx=ln(x)+c doesn't hold?

Of course $\displaystyle \int \frac{1}{x}\,dx=\ln|x|+C$ always holds.

What doesn't hold in general is $\displaystyle \int \frac{1}{f(x)}\,dx=\ln|f(x)|+C$

 

[e^(i Pi)+1=0]

Do you know of the chain rule?

Yes, I meant I don't know trig substitution.

 

[e^(i Pi)+1=0]

Of course $\displaystyle \int \frac{1}{x}\,dx=\ln|x|+C$ always holds.

What doesn't hold in general is $\displaystyle \int \frac{1}{f(x)}\,dx=\ln|f(x)|+C$

Could you provide a few examples please?

 

[Mark44]

You did this problem the hard way, by splitting in into two integrals. An easy substitution will do the trick.

Let u = x² + 2x, so du = (2x + 2)dx
Notice any similarity to the numerator of your original integral?

 

[Dick]

Yes, I meant I don't know trig substitution.

It's not a trig substitution. It's a u substitution. Try putting u=x^2+2x. What is du?

 

[e^(i Pi)+1=0]

To the above two posts -

I know it's not trig substitution, I only mentioned I don't know that because it was mentioned by someone else. U-sub is easy, but I was trying to solve it this way because I'm trying to get a handle on derivatives and integrals of ln(x)

 

[Dick]

To the above two posts -

I know it's not trig substitution, I only mentioned I don't know that because it was mentioned by someone else. U-sub is easy, but I was trying to solve it this way because I'm trying to get a handle on derivatives and integrals of ln(x)

If you don't want to do it the easy way using U-sub, then you'll need to do partial fractions on the second integral. Do you know that?

 

[Bohrok]

If you don't want to do it the easy way using U-sub, then you'll need to do partial fractions on the second integral.

Or complete the square and use a trig substitution?

 

[Dick]

Or complete the square and use a trig substitution?

If doing it the hard way isn't enough and you want to do it the REALLY hard way, that should work.

 

[SammyS]

Of course $\displaystyle \int \frac{1}{x}\,dx=\ln|x|+C$ always holds.

What doesn't hold in general is $\displaystyle \int \frac{1}{f(x)}\,dx=\ln|f(x)|+C$

Could you provide a few examples please?

For instance, if f(x) = x², then

$\displaystyle \int \frac{1}{f(x)}\,dx=\int\frac{1}{x^2}\,dx=\int x^{-2}\,dx=-x^{-1}+C\ne\ln(x^2)+C$

 

[Mentallic]

Could you provide a few examples please?

$$\int\frac{1}{x}dx=\ln|x|+c$$ only holds because $$\frac{d}{dx}\left(\ln|x|+c\right)$$$$=\frac{1}{x}\cdot \frac{d}{dx}(x)=\frac{1}{x}$$ by use of the chain rule, while, in general, $$\frac{d}{dx}\left(\ln|f(x)|+c\right)$$$$=\frac{1}{f(x)}\cdot\frac{d}{dx}f(x)=\frac{f'(x)}{f(x)}$$

So if you have some f(x) in the denominator that needs integrating, unless the numerator is the derivative of the denomintor (or at least a constant multiplied by the derivative) then you can't take the log.