# How do you find #\int _ { 4} ^ { 9} \frac { x + 1} { x+ 2\sqrt { x } - 3} d x#?

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The answer is

Calculate the indefinite integral first

Perform the decomposition into partial fractions

The denominators are the same, compare the numerators

Therefore,

And finally

And the definite integral is

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To solve the integral ( \int_{4}^{9} \frac{x + 1}{x + 2\sqrt{x} - 3} , dx ), we can start by performing a substitution to simplify the integral. Let ( u = \sqrt{x} ), then ( du = \frac{1}{2\sqrt{x}} , dx ). Rearranging, we have ( dx = 2u , du ).

Substituting ( u = \sqrt{x} ) and ( dx = 2u , du ), we get:

[ \int \frac{x + 1}{x + 2\sqrt{x} - 3} , dx = \int \frac{(u^2 + 1) \cdot 2u}{(u^2 + 2u - 3) \cdot 2u} , du ] [ = \int \frac{2u(u^2 + 1)}{2u(u + 3)(u - 1)} , du ] [ = \int \frac{u^2 + 1}{(u + 3)(u - 1)} , du ]

Now we use partial fraction decomposition to split the integrand:

[ \frac{u^2 + 1}{(u + 3)(u - 1)} = \frac{A}{u + 3} + \frac{B}{u - 1} ]

Multiplying both sides by ( (u + 3)(u - 1) ), we get:

[ u^2 + 1 = A(u - 1) + B(u + 3) ]

We solve for ( A ) and ( B ):

[ u^2 + 1 = Au - A + Bu + 3B ]

[ u^2 + 1 = (A + B)u + (-A + 3B) ]

Comparing coefficients, we have:

[ A + B = 0 ] [ -A + 3B = 1 ]

Solving these equations, we find ( A = -\frac{1}{4} ) and ( B = \frac{1}{4} ).

Now, we can rewrite the integral as:

[ \int \left( \frac{-\frac{1}{4}}{u + 3} + \frac{\frac{1}{4}}{u - 1} \right) , du ]

[ = -\frac{1}{4} \ln|u + 3| + \frac{1}{4} \ln|u - 1| + C ]

Finally, substitute back ( u = \sqrt{x} ) and ( dx = 2u , du ):

[ = -\frac{1}{4} \ln|\sqrt{x} + 3| + \frac{1}{4} \ln|\sqrt{x} - 1| + C ]

Now, evaluate this expression from ( x = 4 ) to ( x = 9 ) and subtract the lower limit result from the upper limit result to get the final answer.

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