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Q 20  Integrate the functions \frac{x +2 }{\sqrt { 4x - x ^ 2 }}

Answers (1)

best_answer

let 
x+2 = A\frac{d}{dx}(4x-x^2)+B = A(4-2x)+B
By equating the coefficients and constant term on both sides  we get

A = -1/2 and B = 4

(x+2) = -1/2(4-2x)+4

\\\therefore \int \frac{x+2}{\sqrt{4x-x^2}}dx = -\frac{1}{2}\int \frac{4-2x}{\sqrt{4x-x^2}}+4\int \frac{dx}{\sqrt{4x-x^2}}\\ \ I =\frac{-1}{2}I_1+4I_2....................(i)

Considering I_1
\int \frac{4-2x}{\sqrt{4x-x^2}}dx
let 4x-x^2 =t \Rightarrow (4-2x)dx =dt
I_1=\int \frac{dt}{\sqrt{t}} = 2\sqrt{t}=2\sqrt{4x-x^2}
now, I_2

I_2 =\int \frac{dx}{\sqrt{4x-x^2}} = \int \frac{dx}{\sqrt{2^2-(x-2)^2}}
                                         =\sin^{-1}(\frac{x-2}{2})

put the value of I_1and I_2

I =-\sqrt{4x-x^2}+4\sin^{-1}(\frac{x-2}{2})+C

 

Posted by

manish

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