intergrate (x^2-2x+4)/x

**dangerous_dave** · May 11th 2008, 07:01 PM

How do I intergrate:

(x^2-2x+4)/x

I know it should be easy, can I be pointed in the direction of some information on this please?

**Hibijibi** · May 11th 2008, 07:06 PM

Originally Posted by dangerous_dave

How do I intergrate:

(x^2-2x+4)/x

I know it should be easy, can I be pointed in the direction of some information on this please?

Break it up into x^2/x - 2x/x + 4/x.

This will give you x - 2 + 4/x. I think you can integrate that?

**Mathstud28** · May 11th 2008, 07:06 PM

Originally Posted by dangerous_dave

How do I intergrate:

(x^2-2x+4)/x

I know it should be easy, can I be pointed in the direction of some information on this please?

$\int\frac{x^2-2x+4}{x}dx=\int\bigg[x-2+\frac{4}{x}\bigg]dx=\frac{x^2}{2}-2x+4\ln(x)+C$

**dangerous_dave** · May 11th 2008, 07:29 PM

Oh that does make it easy

. Thanks for your help.

**Mathstud28** · May 11th 2008, 07:32 PM

Originally Posted by dangerous_dave

Oh that does make it easy

. Thanks for your help.

Don't worry we all miss the easy things once in a while

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