Question

Integrate 6x+7/root(x-5)(x-4)

Answer 1

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Answer 2

Answer:

$\int \frac{6x + 7}{ \sqrt{(x - 5)(x - 4)} } dx = \\ \\ 6\sqrt{{x}^{2} - 9x + 20} +34 log( | x-\frac{9}{2}+\sqrt{(x-5)(x-4))} | ) + C$

Step-by-step explanation:

For such type of integration follow the process shown below

$\int \frac{6x + 7}{ \sqrt{(x - 5)(x - 4)} } dx \\ \\ \int \frac{6x + 7}{ \sqrt{ {x}^{2} - 9x + 20 } } dx \\ \\ \\ let \\ \\ 6x + 7 = A \frac{d( {x}^{2} - 9x + 20)}{dx} + B \\ \\ 6x + 7 = A(2x - 9) + B\\ \\ 6x + 7 = 2Ax - 9A + B$

compare the coefficients

$2A = 6 \\ \\ A= 3 \\ \\ - 9A+ B = 7 \\ \\ - 27 + B = 7 \\ \\ B = 34$

$\int \frac{6x + 7}{ \sqrt{ {x}^{2} - 9x + 20 } } dx = \int \frac{3(2x - 9) + 34}{ \sqrt{ {x}^{2} - 9x + 20 } } \\ \\ = 3\int \frac{2x - 9}{ \sqrt{ {x}^{2} - 9x + 20} }dx + \int \frac{34}{\sqrt{ {x}^{2} - 9x + 20}} dx$

Now let

${x}^{2} - 9x + 20 = t \\ \\( 2x - 9)dx = dt$

So

$\int \frac{3(2x - 9) }{ \sqrt{ {x}^{2} - 9x + 20 } } = 3\int \frac{dt}{ \sqrt{t} } \\ \\ = 6\sqrt{t}+ C \\ \\ = 6\sqrt{{x}^{2} - 9x + 20}+ C$

Now convert denominator into complete square method

${x}^{2} - 2x( \frac{9}{2} ) + \frac{81}{4} + 20 - \frac{81}{4} \\ \\ ( {x - \frac{9}{2} )}^{2} - {( \frac{1}{2} )}^{2}$

Integration of

$\int \frac{1}{ \sqrt{ {x}^{2} - {a}^{2} } } dx = log( | x+\sqrt{(x^2-a^2)} | ) + C$

So for

$\int \frac{34}{\sqrt{ {x}^{2} - 9x + 20}} dx \\ \\ 34 \int \frac{1}{ \sqrt{ {(x - \frac{9}{2} )}^{2} - { (\frac{1}{2} )}^{2} } } dx = 34 \: log( | x-\frac{9}{2}-\sqrt{(x^2-9x+20)} | ) + c \\ \\ = 34 \: log( | x-\frac{9}{2}+\sqrt{(x-5)(x-4))} | ) + C \\ \\ \:$

So

$\int \frac{6x + 7}{ \sqrt{(x - 5)(x - 4)} } dx = 6\sqrt{{x}^{2} - 9x + 20} +34 \: log( | x-\frac{9}{2}+\sqrt{(x-5)(x-4))} | ) + C$

Hope it helps you.