- #1

laker88116

- 57

- 0

I have no idea where to start on this. I can't see any substitutions that would work. I tried completing the square. I also tried to split up the fraction. It isn't getting any simpler. Any help is appreciated.

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- Thread starter laker88116
- Start date

- #1

laker88116

- 57

- 0

I have no idea where to start on this. I can't see any substitutions that would work. I tried completing the square. I also tried to split up the fraction. It isn't getting any simpler. Any help is appreciated.

- #2

TD

Homework Helper

- 1,022

- 0

First, calculate the derivative of the denominator and adjust your nominator to get this and split the rest.

[tex]\left( {x^2 + 2x + 5} \right)^\prime = 2x + 2[/tex]

The integral then becomes

[tex]\int {\frac{{x + 4}}{{x^2 + 2x + 5}}} dx = \frac{1}{2}\int {\frac{{2x + 2 + 6}}{{x^2 + 2x + 5}}} dx = \frac{1}{2}\int {\frac{{2x + 2}}{{x^2 + 2x + 5}}} dx + 3\int {\frac{{dx}}{{x^2 + 2x + 5}}}[/tex]

Now the first integral simply becomes the ln of the denominator, because that's how we 'made' it, it's now of the form

The second one no longer has an x in the nominator and you can complete a square in the denominator to get an arctan.

- #3

laker88116

- 57

- 0

I understand that and it makes sense. I just don't get how you determine that's what you do.

- #4

TD

Homework Helper

- 1,022

- 0

When the degree of the nominator isn't lower that the one of the denominator, you can divide first (euclidean division on the polynomials) to get this case again.

- #5

laker88116

- 57

- 0

Alright, I think I got this better now. Thanks.

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