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$\int_0^4 \frac{x+2}{\sqrt{4 x-x^2}} d x=$
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Verified Answer
The correct answer is:
$4 \pi$
$\int \frac{x+2}{\sqrt{4 x-x^2}} d x$
$=\int \frac{(x-2)+4}{\sqrt{4-(x-2)^2}} d x$
$\Rightarrow \quad \int \frac{(x-2)}{\sqrt{4-(x-2)^2}} d x+4 \int \frac{1}{\sqrt{4-(x-2)^2}} d x$
$\Rightarrow \sqrt{4 x-x^2}+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+c$
$\left.\Rightarrow \sqrt{4 x-x^2}-4 \sin ^{-1}\left(\frac{x-2}{2}\right)\right]_0^4$
$\Rightarrow \quad 0+4 \sin ^{-1}(1)-0-4 \sin ^{-1}(-1)$
$\frac{4 \pi}{2}+\frac{4 \pi}{2}=4 \pi$
$=\int \frac{(x-2)+4}{\sqrt{4-(x-2)^2}} d x$
$\Rightarrow \quad \int \frac{(x-2)}{\sqrt{4-(x-2)^2}} d x+4 \int \frac{1}{\sqrt{4-(x-2)^2}} d x$
$\Rightarrow \sqrt{4 x-x^2}+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+c$
$\left.\Rightarrow \sqrt{4 x-x^2}-4 \sin ^{-1}\left(\frac{x-2}{2}\right)\right]_0^4$
$\Rightarrow \quad 0+4 \sin ^{-1}(1)-0-4 \sin ^{-1}(-1)$
$\frac{4 \pi}{2}+\frac{4 \pi}{2}=4 \pi$
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