1. ## integration

is the answer to this indefinite integral $\int\frac{3x^2 - 3x + 10}{(x^4 +4)(x-2)}$ equal to $3(\frac{3}{4}ln{(x-2)} + \frac{1}{4}\ln{(x+2)})$ ?

Also, I've used Partial fractions as a method to integrate. Is this the correct choice?

2. Originally Posted by tsal15
is the answer to this indefinite integral $\int\frac{3x^2 - 3x + 10}{(x^4 +4)(x-2)}$ equal to $3(\frac{3}{4}ln{(x-2)} + \frac{1}{4}\ln{(x+2)})$ ?

Also, I've used Partial fractions as a method to integrate. Is this the correct choice?
If you differentiate your answer you'll see that, unfortunately, you're not even warm.

The partial fraction decomposition you used is almost certainly incorrect. Did you factorise $x^4 + 4$ into two irreducible quadratics ....

Note that:

$x^4 + 4 = x^4 + 4x^2 - 4x^2 + 4 = (x^4 + 4x^2 + 4) - 4x^2 = (x^2 + 2)^2 - 4x^2$

$= (x^2 + 2 - 2x)(x^2 + 2 + 2x) = (x^2 - 2x + 2)(x^2 + 2x + 2)$.

3. Originally Posted by tsal15
is the answer to this indefinite integral $\int\frac{3x^2 - 3x + 10}{(x^4 +4)(x-2)}$ equal to $3(\frac{3}{4}ln{(x-2)} + \frac{1}{4}\ln{(x+2)})$ ?

Also, I've used Partial fractions as a method to integrate. Is this the correct choice?
Attached is what the Wolfram on-line integrator gives as the answer.

4. ohk i think my original post was incorrect...i believe it should be $x^2$ and not $x^4$ so in regards to this correction am i now anywhere near the answer?

Also you're attached picture didn't come out too well (no offence) but I couldn't see anything. so i went to the wolfram site and plugged in the corrected question and i'm still wrong... the wolfram site doesn't show step - by - step on how to do this question. Could you show me how, oh mighty Mr. Fantastic ?

5. Originally Posted by tsal15
ohk i think my original post was incorrect...i believe it should be $x^2$ and not $x^4$ so in regards to this correction am i now anywhere near the answer? Mr F says: Differentiate your answer and see.

Also you're attached picture didn't come out too well (no offence) but I couldn't see anything. Mr F says: Click on the attachment and then click again and you will see it fine.

so i went to the wolfram site and plugged in the corrected question and i'm still wrong... Mr F says: That answers your first question then.

the wolfram site doesn't show step - by - step on how to do this question. Could you show me how, oh mighty Mr. Fantastic ?
It would be much better if you showed all your working. Then what you've done can be reviewed and the errors pointed out.

Did you seek a partial fraction decomposition of the form $\frac{Ax + B}{x^2 + 4} + \frac{C}{x - 2}$ ....?

6. Originally Posted by mr fantastic
It would be much better if you showed all your working. Then what you've done can be reviewed and the errors pointed out.

Did you seek a partial fraction decomposition of the form $\frac{Ax + B}{x^2 + 4} + \frac{C}{x - 2}$ ....?
I used $\frac{A}{x-2} + \frac{Bx + C}{x^2+4}$

So, in the end after using partial fractions i got:

A = 2, B = 1, C = -1

Thus, $\int\frac{2}{x-2} + \frac{x-1}{x^2 + 4} .dx$

and this is where i get stuck. can u show me a few steps that follow?

Thank you Mr. F

7. Originally Posted by tsal15
I used $\frac{A}{x-2} + \frac{Bx + C}{x^2+4}$

So, in the end after using partial fractions i got:

A = 2, B = 1, C = -1

Thus, $\int\frac{2}{x-2} + \frac{x-1}{x^2 + 4} .dx$

and this is where i get stuck. can u show me a few steps that follow?

Thank you Mr. F
Integrating the first term is trivial.

To integrate the second term, note that:

$\frac{x-1}{x^2 + 4} = \frac{x}{x^2 + 4} - \frac{1}{x^2 + 4} = \frac{1}{2} \, \left( \frac{2x}{x^2 + 4} - \frac{2}{x^2 + 2^2}\right)$

8. Originally Posted by mr fantastic
To integrate the second term, note that:

$\frac{x-1}{x^2 + 4} = \frac{x}{x^2 + 4} - \frac{1}{x^2 + 4} = \frac{1}{2} \, \left( \frac{2x}{x^2 + 4} - \frac{2}{x^2 + 2^2}\right)$
Sorry, forgive my ignorance, but how is that useful?

9. Originally Posted by mr fantastic
Integrating the first term is trivial.

To integrate the second term, note that:

$\frac{x-1}{x^2 + 4} = \frac{x}{x^2 + 4} - \frac{1}{x^2 + 4} = \frac{1}{2} \, \left( \frac{2x}{x^2 + 4} - \frac{2}{x^2 + 2^2}\right)$
Originally Posted by tsal15
Sorry, forgive my ignorance, but how is that useful?
If your attempting a question like this there's an expectation that you're familiar with standard integration techniques and results.

Integrate the first term by making the substitution $u = x^2 + 4$.

Integrate the second term by recognising a standard form - you get $\tan^{-1}\left( \frac{x}{2} \right)$.

10. Originally Posted by mr fantastic
If your attempting a question like this there's an expectation that you're familiar with standard integration techniques and results.

Well this is why I'm asking for help on this forum - because my familiarity with the fundamentals isn't to par. And although I am doing my studies of mathematics on my own (it doesn't help when I don't have a teacher to guide me), I'm still tackling everything to the best of my ability too- i assure you

Integrate the first term by making the substitution $u = x^2 + 4$.

Can you please briefly explain, why the substitution of u is necessary? I've realised that the derivative of $ln{(x^2 + 4)} = \frac{2x}{(x^2 + 4)}$...

Integrate the second term by recognising a standard form - you get $\tan^{-1}\left( \frac{x}{2} \right)$.

Got this bit
Thank you

11. Originally Posted by mr fantastic
If your attempting a question like this there's an expectation that you're familiar with standard integration techniques and results.
Well this is why I'm asking for help on this forum - because my familiarity with the fundamentals isn't to par. And although I am doing my studies of mathematics on my own (it doesn't help when I don't have a teacher to guide me), I'm still tackling everything to the best of my ability too- i assure you

Integrate the first term by making the substitution $u = x^2 + 4$.
Can you please briefly explain, why the substitution of u is necessary? I've realised that the derivative of $ln{(x^2 + 4)} = \frac{2x}{(x^2 + 4)}$...

[snip]

Originally Posted by tsal15
Thank you
You need to use the following:

$\int \frac{f'(x)}{f(x)} \, dx = \ln ( f(x) ) + C$.

This is easily proved by making the substitution $u = f(x)$. The reason for the substitution should be obvious in hindsight after differentiating $\frac{1}{f(x)}$ with respect to x.

This is a special case of the more general result $\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du + C$.