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Evaluate the following:
∫⁴₂[{√In(9-x)}/{√In(9-x) + √In(x+3)}]dx​

Evaluate The Following In9xIn9x Inx3dx class=

Sagot :

SalmonWorldTopper

Step-by-step explanation:

[tex] \bf \underline{Given \:Question} \\ [/tex]

Evaluate the following

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

[tex] \red{\large\underline{\sf{Solution-}}}[/tex]

Given integral is

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

Let assume that

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

We know that

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

So, using this property, we get

Change

[tex] \red{\rm :\longmapsto\:x \to \: 2 + 4 - x = 6 - x}[/tex]

[tex]\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[/tex]

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

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

[tex]\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[/tex]

[tex]\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 [/tex]

[tex]\rm :\longmapsto\:2I = \displaystyle\int_{2}^{4}\rm 1dx [/tex]

[tex]\rm :\longmapsto\:2I = \bigg|x\bigg| _{2}^{4}\rm [/tex]

[tex]\rm :\longmapsto\:2I = 4 - 2[/tex]

[tex]\rm :\longmapsto\:2I = 2[/tex]

[tex]\rm :\longmapsto\:I = 1[/tex]

Hence,

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

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Explore more

[tex]\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{a}^{b}\rm f(x) \: dx \: = \: \displaystyle\int_{a}^{b}\rm f(y) \: dy \: }}[/tex]

[tex]\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{a}^{b}\rm f(x) \: dx \: = \: - \: \displaystyle\int_{b}^{a}\rm f(x) \: dx \: }}[/tex]

[tex]\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{0}^{a}\rm f(x) \: dx \: = \: \displaystyle\int_{0}^{a}\rm f(a - x) \: dx \: }}[/tex]

[tex]\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{0}^{2a}\rm f(x) \: dx \: = \:2 \displaystyle\int_{0}^{a}\rm f(x) \: dx \: if \: f(2a - x) = f(x) \: }}[/tex]

[tex]\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{0}^{2a}\rm f(x) \: dx \: = 0 \: if \: f(2a - x) = - f(x) \: }}[/tex]

[tex]\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{ - a}^{a}\rm f(x) \: dx \: = 0 \: if \: f(- x) = - f(x) \: }}[/tex]

[tex]\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{ - a}^{a}\rm f(x) \: dx \: = \:2 \displaystyle\int_{0}^{a}\rm f(x) \: dx \: if \: f( - x) = f(x) \: }}[/tex]

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