Integration with fraction and square root

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SUMMARY

The discussion centers on integrating the function \(\frac{4x+7}{\sqrt{-4x^2+20x-9}}\). A user suggests using the substitution \(u = -4x^2 + 20x - 9\) and its derivative \(du = (-8x + 20)dx\). The integral can be split into two parts: \(\int \frac{4x - 10}{\sqrt{-4x^2 + 20x - 9}}dx\) and \(\int \frac{17}{\sqrt{-4x^2 + 20x - 9}}dx\). The first integral simplifies to \(-\frac{1}{2}\int \frac{du}{u^{1/2}}\), while the second may require a trigonometric substitution.

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i am given this function and need to integrate it

[tex]\frac{4x+7}{\sqrt{-4x^2+20x-9}}[/tex]

i have been trying to intergrate it by calling something T, preferably the expression under the sqrd root, or part thereof (-4x2+20x=T) but i can't find the way to do this, can't find the best expression that fits both the denominator and numerator of the fraction.. any other ideas??
 
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Try u = -4x^2 + 20x - 9, and du = (-8x + 20)dx (I like u for substitutions better than T.)

Your numerator is 4x + 7 = 4x -10 + 17, so your integral can be rewritten as two integrals.

[tex]\int \frac{4x - 10}{\sqrt{-4x^2 + 20x - 9}}dx + \int \frac{17}{\sqrt{-4x^2 + 20x - 9}}dx[/tex]

Applying the substitution in the first integral, we have [itex]-1/2\int du/u^{1/2}[/itex]. The second one is probably amenable to a trig substitution.
 

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