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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals4 Marks
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 - x2 = 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 - x2 = -(-4x + x2)
= (-4x + x2 + 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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