# What is the integral of #(x^2 +2x-1)/(x^2+9)#?

First of all split the fraction up like so:

We can now integrate each fraction one by one, i.e:

We will have to do a bit of rearranging of the first fraction (add and subtract 9):

This can now be rearranged like so:

-And for the third term:

Reverse the substitution and we get:

So returning to our original integral if apply what have found we get that:

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To find the integral of ((x^2 + 2x - 1)/(x^2 + 9)), you can use the method of partial fraction decomposition. After decomposing the fraction into simpler fractions, you can integrate each term individually.

The integral of ((x^2 + 2x - 1)/(x^2 + 9)) can be expressed as: [ \int \frac{x^2 + 2x - 1}{x^2 + 9} ,dx = \int \left(1 - \frac{9}{x^2 + 9}\right) ,dx ]

Then, integrating each term separately: [ \int 1 ,dx = x ] [ \int \frac{9}{x^2 + 9} ,dx = 3 \arctan\left(\frac{x}{3}\right) ]

Therefore, the integral of ((x^2 + 2x - 1)/(x^2 + 9)) is: [ x - 3 \arctan\left(\frac{x}{3}\right) + C ] where (C) is the constant of integration.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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