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# If α = $\int\limits_{0}^{1}{({{e}^{(9x+3{{\tan }^{-1}}x)}})}\left( \dfrac{12+9{{x}^{2}}}{1+{{x}^{2}}} \right)dx,$ where ${{\tan }^{-1}}x$ takes only principal values, then the value of $\left( {{\log }_{e}}|1+a|-\dfrac{3\pi }{4} \right)$ is

Last updated date: 09th Sep 2024
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Hint: Now we have been given with α = $\int\limits_{0}^{1}{({{e}^{(9x+3{{\tan }^{-1}}x)}})}\left( \dfrac{12+9{{x}^{2}}}{1+{{x}^{2}}} \right)dx,$ to solve this integral we will substitute $9x+3{{\tan }^{-1}}x$ as t and solve it by substituting method. Once we find the value of α we will substitute the value of α in $\left( {{\log }_{e}}|1+a|-\dfrac{3\pi }{4} \right)$ and find the solution.

Now consider the integral α = $\int\limits_{0}^{1}{({{e}^{(9x+3{{\tan }^{-1}}x)}})}\left( \dfrac{12+9{{x}^{2}}}{1+{{x}^{2}}} \right)dx,$
Now here we can see that $d({{\tan }^{-1}}x)=\dfrac{1}{1+{{x}^{2}}}$ hence somehow substitution can be used to simplify the problem.
Now let us take $9x+3{{\tan }^{-1}}x=t$ Differentiating on both side with respect to x we get
$9+3\dfrac{1}{1+{{x}^{2}}}=\dfrac{dt}{dx}$
Solving left hand side we get
$\dfrac{9+9{{x}^{2}}+3}{1+{{x}^{2}}}=\dfrac{dt}{dx}$
Hence we have
$\dfrac{12+9{{x}^{2}}}{1+{{x}^{2}}}=\dfrac{dt}{dx}$
Taking dx on left hand side we get
$\left( \dfrac{12+9{{x}^{2}}}{1+{{x}^{2}}} \right)dx=dt$
Similarly let us check the change in limit of integral
As $\displaystyle \lim_{x \to 0}9(0)+{{\tan }^{-1}}0=0$
Similarly as $\displaystyle \lim_{x \to 1}9(1)+3{{\tan }^{-1}}1=9+\dfrac{3\pi }{4}$
Now using this substitution we get in the given integration we get.
\begin{align} & \int\limits_{0}^{1}{({{e}^{(9x+3{{\tan }^{-1}}x)}})}\left( \dfrac{12+9{{x}^{2}}}{1+{{x}^{2}}} \right)dx=\int\limits_{0}^{9+\dfrac{3\pi }{4}}{{{e}^{t}}dt} \\ & =[{{e}^{t}}]_{0}^{^{9+\dfrac{3\pi }{4}}} \\ & ={{e}^{9+\dfrac{3\pi }{4}}}-{{e}^{0}} \\ & ={{e}^{9+\dfrac{3\pi }{4}}}-1 \\ \end{align}
Hence the value of α is equal to ${{e}^{9+\dfrac{3\pi }{4}}}-1$
Now since α = ${{e}^{9+\dfrac{3\pi }{4}}}-1$ adding 1 on both sides we get
α + 1 = ${{e}^{9+\dfrac{3\pi }{4}}}-1+1$
Hence we get the value of α + 1 = ${{e}^{9+\dfrac{3\pi }{4}}}$
Now taking log on both sides we get.
${{\log }_{e}}|a+1|={{\log }_{e}}{{e}^{9+\dfrac{3\pi }{4}}}$
But we know ${{\log }_{e}}{{e}^{a}}=a$ using this we get
${{\log }_{e}}|a+1|=9+\dfrac{3\pi }{4}$
Now let us subtract $\dfrac{3\pi }{4}$ on both sides.
$\left( {{\log }_{e}}|1+a|-\dfrac{3\pi }{4} \right)=9+\dfrac{3\pi }{4}-\dfrac{3\pi }{4}=9$

Hence we get the value of $\left( {{\log }_{e}}|1+a|-\dfrac{3\pi }{4} \right)$ = 9.

Note: Here when we use a method of substitution to integrate note that the limits of integration also change. Hence if we substitute a function f(x) as t we should change the limits of x to t by substituting the value of x in substitution. Also since we are given ${{\tan }^{-1}}x$ takes only principal values we could write ${{\tan }^{-1}}1=\dfrac{\pi }{4},{{\tan }^{-1}}0=0$