Integrate the function x+2 4 x-x^ 2 - Vidyadip

Question

Integrate the function $\frac{x+2}{\sqrt{4 x-x^{2}}}$

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Answer

Let $x + 2 = A \frac{d}{d x}\left(4 x-x^{2}\right)+B$
$\Rightarrow x + 2 = A(4 -2x) + B$
Now, equating the coefficients of $x$ and constant term on both sides, we get,
$-2A = 1$
$\Rightarrow A=-\frac{1}{2}$
and $4A + B = 2$
$\Rightarrow B = 4$
$\Rightarrow x+2=-\frac{1}{2}(4-2 x)+4$
Now, $\int \frac{x+2}{\sqrt{4 x-x^{2}}} d x=\int \frac{-\frac{1}{2}(4-2 x)+4}{\sqrt{4 x-x^{2}}} d x$
$\Rightarrow -\frac{1}{2} \int \frac{(4-2 x)}{\sqrt{4 x-x^{2}}} d x+4 \int \frac{1}{\sqrt{4 x-x^{2}}} d x$
Now, let us consider, $\int \frac{(4-2 x)}{\sqrt{4 x-x^{2}}} d x$
Let $4x - x^2 = t$
$\Rightarrow (4 - 2x) dx = dt$
$\therefore \int \frac{(4-2 x)}{\sqrt{4 x-x^{2}}} d x=\int \frac{d t}{\sqrt{t}}=2 \sqrt{t}=2 \sqrt{4 x-x^{2}}.......(i)$
And, Now let us consider, $\int \frac{1}{\sqrt{4 x-x^{2}}} d x$
Then, $4x - x^2 = -(-4x + x^2)$
$= (-4x + x^2 + 4 - 4)$
$= 4 - (x - 2)^{2 }$
$= (2)^2 - (x - 2)^{2 }$
$\therefore \int \frac{1}{\sqrt{4 x-x^{2}}} d x=\sin ^{-1}\left(\frac{x-2}{2}\right) ......(ii)$
using eq. $(i)$ and $(ii),$ we get,
$\Rightarrow \int \frac{x+2}{\sqrt{4 x-x^{2}}} d x=-\sqrt{4 x-x^{2}}+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+C$

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