# How do you find the definite integral for: #(2+3x) / (4+x2) dx# for the intervals #[0, 1]#?

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To find the definite integral of ( \frac{2 + 3x}{4 + x^2} ) with respect to ( x ) over the interval ([0, 1]), follow these steps:

- Split the fraction into partial fractions if possible.
- Integrate each term separately.
- Evaluate the definite integral using the Fundamental Theorem of Calculus.

If we split ( \frac{2 + 3x}{4 + x^2} ) into partial fractions, we get:

[ \frac{2 + 3x}{4 + x^2} = \frac{A}{2 + x} + \frac{B}{2 - x} ]

Solve for ( A ) and ( B ), and integrate each term:

[ A = \frac{5}{4}, \quad B = \frac{3}{4} ]

The integral of ( \frac{5}{4(2 + x)} ) with respect to ( x ) is ( \frac{5}{4} \ln|2 + x| + C_1 ). The integral of ( \frac{3}{4(2 - x)} ) with respect to ( x ) is ( -\frac{3}{4} \ln|2 - x| + C_2 ).

Now, evaluate each integral from ( x = 0 ) to ( x = 1 ):

[ \int_{0}^{1} \frac{2 + 3x}{4 + x^2} , dx = \left( \frac{5}{4} \ln|2 + x| - \frac{3}{4} \ln|2 - x| \right) \bigg|_{0}^{1} ]

[ = \left( \frac{5}{4} \ln|3| - \frac{3}{4} \ln|1| \right) - \left( \frac{5}{4} \ln|2| - \frac{3}{4} \ln|2| \right) ]

[ = \frac{5}{4} \ln\left(\frac{3}{2}\right) - \frac{3}{4} \ln(1) - \frac{5}{4} \ln(2) + \frac{3}{4} \ln(2) ]

[ = \frac{5}{4} \ln\left(\frac{3}{2}\right) - \frac{5}{4} \ln(2) ]

[ = \frac{5}{4} \ln\left(\frac{3}{4}\right) ]

Therefore, the definite integral of ( \frac{2 + 3x}{4 + x^2} ) over the interval ([0, 1]) is ( \frac{5}{4} \ln\left(\frac{3}{4}\right) ).

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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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