Question

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Answer 1

Given Question

Evaluate the following

$\displaystyle\int_{2}^{4}\rm \frac{ \sqrt{ln(9 - x)} }{ \sqrt{ln(9 - x)} + \sqrt{ln(x + 3)} } \: dx$

$\red{\large\underline{\sf{Solution-}}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int_{2}^{4}\rm \frac{ \sqrt{ln(9 - x)} }{ \sqrt{ln(9 - x)} + \sqrt{ln(x + 3)} } \: dx$

Let assume that

$\rm :\longmapsto\:I = \displaystyle\int_{2}^{4}\rm \frac{ \sqrt{ln(9 - x)} }{ \sqrt{ln(9 - x)} + \sqrt{ln(x + 3)} } \: dx - - - (1)$

We know that

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{a}^{b}\rm f(x) \: dx \: = \: \displaystyle\int_{a}^{b}\rm f(a + b - x) \: dx \: }}$

So, using this property, we get

Change

$\red{\rm :\longmapsto\:x \to \: 2 + 4 - x = 6 - x}$

$\rm :\longmapsto\:I = \displaystyle\int_{2}^{4}\rm \frac{ \sqrt{ln(9 - (6 - x))} }{ \sqrt{ln(9 -(6 - x))} + \sqrt{ln(6 - x + 3)} } \: dx$

$\rm :\longmapsto\:I = \displaystyle\int_{2}^{4}\rm \frac{ \sqrt{ln(3 + x)} }{ \sqrt{ln(3 + x)} + \sqrt{ln(9 - x)} } \: dx - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\:2I = \displaystyle\int_{2}^{4}\rm \frac{ \sqrt{ln(9 - x)} }{ \sqrt{ln(9 - x)} + \sqrt{ln(x + 3)}}dx + \displaystyle\int_{2}^{4}\rm \frac{ \sqrt{ln(x + 3)} }{ \sqrt{ln(9 - x)} + \sqrt{ln(x + 3)}}dx$

$\rm :\longmapsto\:2I = \displaystyle\int_{2}^{4}\rm \frac{ \sqrt{ln(9 - x)} + ln \sqrt{(x + 3)} }{ \sqrt{ln(9 - x)} + \sqrt{ln(x + 3)}}dx$

$\rm :\longmapsto\:2I = \displaystyle\int_{2}^{4}\rm 1dx$

$\rm :\longmapsto\:2I = \bigg|x\bigg| _{2}^{4}\rm$

$\rm :\longmapsto\:2I = 4 - 2$

$\rm :\longmapsto\:2I = 2$

$\rm :\longmapsto\:I = 1$

Hence,

$\boxed{\tt{ \displaystyle\int_{2}^{4}\rm \frac{ \sqrt{ln(9 - x)} }{ \sqrt{ln(9 - x)} + \sqrt{ln(x + 3)} } \: dx = 1}}$

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