# How do you integrate #1/(x^2+x) dx#?

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To integrate ( \frac{1}{x^2 + x} ) with respect to ( x ), you can use partial fraction decomposition.

Here's how:

- Rewrite the integrand as ( \frac{1}{x(x + 1)} ).
- Decompose ( \frac{1}{x(x + 1)} ) into partial fractions: ( \frac{A}{x} + \frac{B}{x + 1} ).
- Find the values of ( A ) and ( B ) by equating numerators: ( 1 = A(x + 1) + Bx ).
- Solve for ( A ) and ( B ).
- Once you have ( A ) and ( B ), integrate each term separately.
- The integral of ( \frac{A}{x} ) is ( A\ln|x| ), and the integral of ( \frac{B}{x + 1} ) is ( B\ln|x + 1| ).
- Combine the results to get the final answer.

The integration will yield:

[ \int \frac{1}{x^2 + x} , dx = \ln|x| - \ln|x + 1| + 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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