Question

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

Answer 1

Hint: Now we have been given with α = $\underset{0}{\overset{1}{\int }}(e^{(9x+3{\tan }^{-1}x)})\left({\displaystyle \frac{12+9x^2}{1+x^2}}\right)dx,$ 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 $({\log }_e|1+a|-{\displaystyle \frac{3\pi }{4}})$ and find the solution.

Complete step by step answer:
Now consider the integral α = $\underset{0}{\overset{1}{\int }}(e^{(9x+3{\tan }^{-1}x)})\left({\displaystyle \frac{12+9x^2}{1+x^2}}\right)dx,$
Now here we can see that $d({\tan }^{-1}x)={\displaystyle \frac{1}{1+x^2}}$ 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
$9+3{\displaystyle \frac{1}{1+x^2}}={\displaystyle \frac{dt}{dx}}$
Solving left hand side we get
${\displaystyle \frac{9+9x^2+3}{1+x^2}}={\displaystyle \frac{dt}{dx}}$
Hence we have
${\displaystyle \frac{12+9x^2}{1+x^2}}={\displaystyle \frac{dt}{dx}}$
Taking dx on left hand side we get
$\left({\displaystyle \frac{12+9x^2}{1+x^2}}\right)dx=dt$
Similarly let us check the change in limit of integral
As ${\displaystyle \underset{x\to 0}{lim}9(0)+{\tan }^{-1}0=0}$
Similarly as ${\displaystyle \underset{x\to 1}{lim}9(1)+3{\tan }^{-1}1=9+{\displaystyle \frac{3\pi }{4}}}$
Now using this substitution we get in the given integration we get.
$\begin{array}{rl}& \underset{0}{\overset{1}{\int }}(e^{(9x+3{\tan }^{-1}x)})\left({\displaystyle \frac{12+9x^2}{1+x^2}}\right)dx=\underset{0}{\overset{9+{\displaystyle \frac{3\pi }{4}}}{\int }}{e}^{t}dt\\ & =[e^t{]}_0^{{}^{9+{\displaystyle \frac{3\pi }{4}}}}\\ & =e^{9+{\displaystyle \frac{3\pi }{4}}}-e^0\\ & =e^{9+{\displaystyle \frac{3\pi }{4}}}-1\end{array}$
Hence the value of α is equal to $e^{9+{\displaystyle \frac{3\pi }{4}}}-1$
Now since α = $e^{9+{\displaystyle \frac{3\pi }{4}}}-1$ adding 1 on both sides we get
α + 1 = $e^{9+{\displaystyle \frac{3\pi }{4}}}-1+1$
Hence we get the value of α + 1 = $$e^{9+{\displaystyle \frac{3\pi }{4}}}$$
Now taking log on both sides we get.
${\log }_e|a+1|={\log }_e{e}^{9+{\displaystyle \frac{3\pi }{4}}}$
But we know ${\log }_e{e}^a=a$ using this we get
${\log }_e|a+1|=9+{\displaystyle \frac{3\pi }{4}}$
Now let us subtract ${\displaystyle \frac{3\pi }{4}}$ on both sides.
$$({\log }_e|1+a|-{\displaystyle \frac{3\pi }{4}})=9+{\displaystyle \frac{3\pi }{4}}-{\displaystyle \frac{3\pi }{4}}=9$$

Hence we get the value of $({\log }_e|1+a|-{\displaystyle \frac{3\pi }{4}})$ = 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 ${\tan }^{-1}1={\displaystyle \frac{\pi }{4}},{\tan }^{-1}0=0$