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Q19   Integrate the functions \frac{6x + 7 }{\sqrt {( x-5 )( x-4)}}

Answers (1)

best_answer

let 
6x+7 = A\frac{d}{dx}(x^2-9x+20)+B =A(2x-9)+B
By comparing the coefficients and constants on both sides, we get
A =3  and B =34

I =\int \frac{6x+7}{\sqrt{x^2-9x+20}}dx = \int \frac{3(2x+9)}{\sqrt{x^2-9x+20}}dx+34\int\frac{dx}{\sqrt{x^2-9x+20}}I = I_1+I_2....................................(i)

Considering I_1

I_1 =\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx        let x^2-9x+20 = t \Rightarrow (2x-9)dx =dt

I_1=\int \frac{dt}{\sqrt{t}} = 2\sqrt{t}=2\sqrt{x^2-9x+20}

Now consider I_2

I_2=\int \frac{dx}{\sqrt{x^2-9x+20}}
here the denominator can be also written as 
Dr = (x-\frac{9}{2})^2-(\frac{1}{2})^2

\therefore I_2 = \int \frac{dx}{\sqrt{(x-\frac{9}{2})^2-(\frac{1}{2})^2}}
           \\= \log\left | (x-\frac{9}{2})^2+\sqrt{x^2-9x+20} \right |

Now put the values of I_1 and I_2 in eq (i)

\\I = 3I_1+34I_2\\ I=6\sqrt{x^2-9x+20}+34\log\left | (x-\frac{9}{2})+\sqrt{x^2-9x+20} \right |+C

Posted by

manish

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