The integral $$ \int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) d x $$ is equal to :

To solve the integral: $ I = \int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) \, dx $ we start by simplifying the logarithmic term: $ \ln\left(\frac{e x}{2}\right) = \ln(e) + \ln\left(\frac{x}{2}\right) = 1 + \ln\left(\frac{x}{2}\right) $ So the integral becomes: $ I = \int \left[ \left( \frac{x}{2} \right)^x + \left( \frac{2}{x} \right)^x \right] \left[ 1 + \ln\left( \frac{x}{2} \right) \right] \, dx $ Let's denote: $ A = \left( \dfrac{x}{2}

Indefinite Integration — Page 5 | JEE Mathematics

Q41

For

\alpha, \beta, \gamma, \delta \in \mathbb{N}

, if

\int\left(\left(\dfrac{x}{e}\right)^{2 x}+\left(\dfrac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\dfrac{1}{\alpha}\left(\dfrac{x}{e}\right)^{\beta x}-\dfrac{1}{\gamma}\left(\dfrac{e}{x}\right)^{\delta x}+C

, where

e=\sum\limits_{n=0}^{\infty} \dfrac{1}{n !}

and

\mathrm{C}

is constant of integration, then

\alpha+2 \beta+3 \gamma-4 \delta

is equal to :

A

-8

B

-4

C 1

D 4

Correct Answer

Option D

Solution

We have,

\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C

\begin{aligned} \left(\frac{x}{e}\right)^{2 x} & =\left(\frac{e^{\log _e x}}{e}\right)^{2 x} \left[\because x=e^{\log _e x}\right] \\\\ & =e^{2\left(x \log _e x-x\right)} .........(i) \end{aligned}

\text{ Also, }\left(\frac{e}{x}\right)^{2 x}=\left(\frac{x}{e}\right)^{-2 x}=e^{-2\left(x \log _e x-x\right)}

.........(ii) Let,

\begin{aligned} I & =\int\left[\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right] \log _e x d x \\\\ & =\int\left[e^{2\left(x \log _e x-x\right)}+e^{-2\left(x \log _e x-x\right)}\right] \log _e x d x ~~~~\text{ [From Eqs. (i) and (ii)] } \end{aligned}

Let $x \log _e x-x=t$

\log _e x d x=d t

\begin{aligned} I & =\int\left(e^{2 t}+e^{-2 t}\right) d t=\frac{e^{2 t}}{2}-\frac{e^{-2 t}}{2}+C \\\\ & =\frac{1}{2}\left(\frac{x}{e}\right)^{2 x}-\frac{1}{2}\left(\frac{e}{x}\right)^{2 x}+C \end{aligned}

On comparing I with $\dfrac{1}{\alpha}\left(\dfrac{x}{e}\right)^{\beta x}-\dfrac{1}{\gamma}\left(\dfrac{e}{x}\right)^{\delta x}+C$

\begin{aligned} & \alpha=2, \beta=2, \gamma=2, \delta=2 \\\\ & \therefore \alpha+2 \beta+3 \gamma-4 \delta=2+2 \times 2+3 \times 2-4 \times 2=4 \end{aligned}

Q42

If

I(x) = \int {{e^{{{\sin }^2}x}}(\cos x\sin 2x - \sin x)dx}

and

I(0) = 1

, then

I\left( {{\pi \over 3}} \right)

is equal to :

A

- {e^{{3 \over 4}}}

B

- {1 \over 2}{e^{{3 \over 4}}}

C

{e^{{3 \over 4}}}

D

{1 \over 2}{e^{{3 \over 4}}}

Correct Answer

Option D

Solution

\begin{aligned} & \text{ Given, } I(x)=\int e^{\sin ^2 x}(\cos x \sin 2 x-\sin x) d x \\\\ & =\int e^{\sin ^2 x} \cdot \cos x \cdot \sin 2 x d x-\int \sin x e^{\sin ^2 x} d x \\\\ & =\int \frac{\cos x}{\mathrm{I}} \cdot \frac{e^{\sin ^2 x} \cdot \sin 2 x}{\mathrm{II}} d x-\int \sin x \cdot e^{\sin ^2 x} d x \\\\ & =\cos x \cdot e^{\sin ^2 x}-\int(-\sin x) e^{\sin ^2 x} d x-\int \sin x e^{\sin ^2 x} d x \\\\ & =\cos x \cdot e^{\sin ^2 x}+\int \sin x e^{\sin ^2 x} \cdot d x-\int \sin x \cdot e^{\sin ^2 x} d x+C \end{aligned}

$I(x)=e^{\sin ^2 x} \cdot \cos x+C$ Given, $I(0)=1 \Rightarrow C=0$ So, $I(x)=e^{\sin ^2 x} \cdot \cos x$

\Rightarrow I(\pi / 3)=e^{3 / 4} \cdot \frac{1}{2}=\frac{e^{3 / 4}}{2}

Q43

The integral

\int\left[\left(\dfrac{x}{2}\right)^x+\left(\dfrac{2}{x}\right)^x\right] \ln \left(\dfrac{e x}{2}\right) d x

is equal to :

A

\left(\dfrac{x}{2}\right)^{x}+\left(\dfrac{2}{x}\right)^{x}+C

B

\left(\dfrac{x}{2}\right)^{x}-\left(\dfrac{2}{x}\right)^{x}+C

C

\left(\dfrac{x}{2}\right)^{x} \log _{2}\left(\dfrac{2}{x}\right)+C

D None

Correct Answer

Option B

Solution

To solve the integral: $I = \int\left[\left(\dfrac{x}{2}\right)^x+\left(\dfrac{2}{x}\right)^x\right] \ln \left(\dfrac{e x}{2}\right) \, dx$ we start by simplifying the logarithmic term: $\ln\left(\dfrac{e x}{2}\right) = \ln(e) + \ln\left(\dfrac{x}{2}\right) = 1 + \ln\left(\dfrac{x}{2}\right)$ So the integral becomes: $I = \int \left[ \left( \dfrac{x}{2} \right)^x + \left( \dfrac{2}{x} \right)^x \right] \left[ 1 + \ln\left( \dfrac{x}{2} \right) \right] \, dx$ Let's denote: $A = \left( \dfrac{x}{2} \right)^x$ $B = \left( \dfrac{2}{x} \right)^x$ Then, the integral can be split: $I = \int \left[ A(1 + \ln\left( \dfrac{x}{2} \right)) + B(1 + \ln\left( \dfrac{x}{2} \right)) \right] \, dx$ Calculating $\int A(1 + \ln\left( \dfrac{x}{2} \right)) \, dx$ : First, find the derivative of $A$ : $\dfrac{dA}{dx} = A \left[ \ln\left( \dfrac{x}{2} \right) + 1 \right]$ This means: $A \left[ 1 + \ln\left( \dfrac{x}{2} \right) \right] = \dfrac{dA}{dx}$ Therefore: $\int A \left[ 1 + \ln\left( \dfrac{x}{2} \right) \right] \, dx = \int \dfrac{dA}{dx} \, dx = A + C_1 = \left( \dfrac{x}{2} \right)^x + C_1$ Calculating $\int B(1 + \ln\left( \dfrac{x}{2} \right)) \, dx$ : Note that: $\ln\left( \dfrac{x}{2} \right) = \ln x - \ln 2$ $\ln\left( \dfrac{2}{x} \right) = \ln 2 - \ln x$ The derivative of $B$ is: $\dfrac{dB}{dx} = B \left[ \ln\left( \dfrac{2}{x} \right) + 1 \right] = B \left[ \ln 2 - \ln x + 1 \right]$ But in our integrand, we have $B \left[ 1 + \ln\left( \dfrac{x}{2} \right) \right] = B \left[ 1 + \ln x - \ln 2 \right]$ .

Notice that: $1 + \ln x - \ln 2 = - \left( \ln 2 - \ln x + 1 \right)$ Thus: $B \left[ 1 + \ln\left( \dfrac{x}{2} \right) \right] = - \dfrac{dB}{dx}$ Therefore: $\int B \left[ 1 + \ln\left( \dfrac{x}{2} \right) \right] \, dx = -\int \dfrac{dB}{dx} \, dx = -B + C_2 = -\left( \dfrac{2}{x} \right)^x + C_2$ Combining the Results: Adding both integrals: $I = \left( \dfrac{x}{2} \right)^x - \left( \dfrac{2}{x} \right)^x + C$ Answer: Option B $\left( \dfrac{x}{2} \right)^x - \left( \dfrac{2}{x} \right)^x + C$

Q44

Let

I(x)=\int \dfrac{(x+1)}{x\left(1+x e^{x}\right)^{2}} d x, x > 0

. If

\lim\limits_{x \rightarrow \infty} I(x)=0

, then

I(1)

is equal to :

A

\dfrac{e+1}{e+2}-\log _{e}(e+1)

B

\dfrac{e+1}{e+2}+\log _{e}(e+1)

C

\dfrac{e+2}{e+1}-\log _{e}(e+1)

D

\dfrac{e+2}{e+1}+\log _{e}(e+1)

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{I}=\int \frac{x+1}{x\left(1+x e^x\right)^2} d x \\\\ & \text{ Put } 1+x e^x=t \Rightarrow x e^x=t-1 \\\\ & \Rightarrow\left(x e^x+e^x\right) d x=d t \\\\ & \Rightarrow e^x(x+1) d x=d t \end{aligned}

\therefore I=\int \frac{d t}{e^x \cdot x t^2}=\int \frac{d t}{(t-1) t^2}

Let $\dfrac{1}{t^2(t-1)}=\dfrac{\mathrm{A}}{(t-1)}+\dfrac{\mathrm{B} t+\mathrm{C}}{t^2}$

\Rightarrow 1=\mathrm{A} t^2+(\mathrm{B} t+\mathrm{C})(t-1)

Comparing coefficients of $t^2, t$ and constant terms, we get

A+B=0, C-B=0,-C=1

On solving above equations, we get

\begin{aligned} & \mathrm{C}=-1,=\mathrm{B}, \mathrm{A}=1 \\\\ & \therefore \mathrm{I}=\int \frac{1}{t-1} d t+\int \frac{-t-1}{t^2} d t \\\\ & =\int \frac{1}{t-1} d t-\int \frac{1}{t} d t-\int \frac{1}{t^2} d t \\\\ & =\log |t-1|-\log |t|+\frac{1}{t}+\mathrm{C} \\\\ & \Rightarrow \mathrm{I}=\log \left|x e^x\right|-\log \left|1+x e^x\right|+\frac{1}{1+x e^x}+c \end{aligned}

=\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x}+C

Now, $\lim _{x \rightarrow \infty} \mathrm{I}(x)=0$

\begin{aligned} & \Rightarrow \lim _{x \rightarrow \infty}\left\{\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x}+\mathrm{C}\right\}=0 \\\\ & \Rightarrow \lim _{x \rightarrow \infty}\left\{\log \left|\frac{e^x}{\frac{1}{x}+e^x}\right|+\frac{\frac{1}{x}}{\frac{1}{x}+e^x}+\mathrm{C}\right\} \\\\ & \Rightarrow 0+0+\mathrm{C}=0 \Rightarrow \mathrm{C}=0 \\\\ & \therefore \mathrm{I}(x)=\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x} \\\\ & \Rightarrow \mathrm{I}(1)=\log \left|\frac{e}{1+e}\right|+\frac{1}{1+e}=1-\log (1+e)+\frac{1}{1+e} \\\\ & =\frac{2+e}{1+e}-\log |1+e| \end{aligned}

Q45

Let

I(x)=\int \dfrac{x^{2}\left(x \sec ^{2} x+\tan x\right)}{(x \tan x+1)^{2}} d x

. If

I(0)=0

, then

I\left(\dfrac{\pi}{4}\right)

is equal to :

A

\log _{e} \dfrac{(\pi+4)^{2}}{32}-\dfrac{\pi^{2}}{4(\pi+4)}

B

\log _{e} \dfrac{(\pi+4)^{2}}{16}-\dfrac{\pi^{2}}{4(\pi+4)}

C

\log _{e} \dfrac{(\pi+4)^{2}}{16}+\dfrac{\pi^{2}}{4(\pi+4)}

D

\log _{e} \dfrac{(\pi+4)^{2}}{32}+\dfrac{\pi^{2}}{4(\pi+4)}

Correct Answer

Option A

Solution

We have,

\begin{aligned} I(x)= & \int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x \\\\ = & x^2 \int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x \\\\ & \quad-\int\left\{\frac{d}{d x}\left(x^2\right) \int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x\right\} d x \text{ (integration by parts) } \end{aligned}

=x^2\left(\frac{-1}{x \tan x+1}\right)+\int \frac{2 x}{x \tan x+1} d x

Now, let

\begin{aligned} I_1 & =2 \int \frac{x}{x \tan x+1} d x \\\\ & =2 \int \frac{x \cos x}{x \sin x+\cos x} d x \end{aligned}

\begin{aligned} & \text{ On putting } x \sin x+\cos x=t \\\\ & \Rightarrow (x \cos x+\sin x-\sin x) d x=d t \\\\ & \Rightarrow x \cos x d x=d t \end{aligned}

\begin{aligned} &\therefore I_1 =2 \int \frac{d t}{t}=2 \log t+c \\\\ & =2 \log (x \sin x+\cos x)+c \end{aligned}

\begin{gathered} \therefore I(x)=\frac{-x^2}{x \tan x+1}+2 \log(x \sin x+\cos x)+c \end{gathered}

When, $x=0$ , then

\begin{array}{ll} & I(0)=0+2 \log (1)+c=0 \\\\ &\Rightarrow c=0 \\\\ &\therefore I(x)=\frac{-x^2}{x \tan x+1}+2 \log (x \sin x+\cos x) \end{array}

\begin{aligned} &\therefore I(x) =\frac{-x^2}{x \tan x+1}+2 \log (x \sin x+\cos x) \\\\ &\Rightarrow I\left(\frac{\pi}{4}\right) =\frac{-\frac{\pi^2}{16}}{\frac{\pi}{4}+1}+2 \log \left(\frac{\pi}{4} \times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) \\\\ & =\log \left(\frac{(\pi+4)^2}{32}\right)-\frac{\pi^2}{4(\pi+4)} \end{aligned}

Q46

\text{ The integral } \int \frac{\left(x^8-x^2\right) \mathrm{d} x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} \text{ is equal to : }

A

\log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\dfrac{1}{x^3}\right)\right|\right)^{1 / 3}+\mathrm{C}

B

\log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\dfrac{1}{x^3}\right)\right|\right)+\mathrm{C}

C

\log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\dfrac{1}{x^3}\right)\right|\right)^{1 / 2}+\mathrm{C}

D

\log _{\mathrm{e}}\left(\left|\tan ^{-1}\left(x^3+\dfrac{1}{x^3}\right)\right|\right)^3+\mathrm{C}

Correct Answer

Option A

Solution

I=\int \frac{x^8-x^2}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} d x

\begin{aligned} & \text{ Let } \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)=t \\ & \Rightarrow \frac{1}{1+\left(x^3+\frac{1}{x^3}\right)^2} \cdot\left(3 x^2-\frac{3}{x^4}\right) d x=d t \\ & \Rightarrow \frac{x^6}{x^{12}+3 x^6+1} \cdot \frac{3 x^6-3}{x^4} d x=d t \\ & I=\frac{1}{3} \int \frac{d t}{t}=\frac{1}{3} \ln |t|+C \\ & I=\frac{1}{3} \ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|+C \\ & I=\ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|^{1 / 3}+C \end{aligned}

Hence option (1) is correct.

Q47

For

x \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)

, if

y(x)=\int \dfrac{\operatorname{cosec} x+\sin x}{\operatorname{cosec} x \sec x+\tan x \sin ^2 x} d x

, and

\lim \limits_{x \rightarrow\left(\dfrac{\pi}{2}\right)^{-}} y(x)=0

then

y\left(\dfrac{\pi}{4}\right)

is equal to

A

-\dfrac{1}{\sqrt{2}} \tan ^{-1}\left(\dfrac{1}{\sqrt{2}}\right)

B

\tan ^{-1}\left(\dfrac{1}{\sqrt{2}}\right)

C

\dfrac{1}{2} \tan ^{-1}\left(\dfrac{1}{\sqrt{2}}\right)

D

\dfrac{1}{\sqrt{2}} \tan ^{-1}\left(-\dfrac{1}{2}\right)

Correct Answer

Option D

Solution

\begin{aligned} & y(x)=\int \frac{\left(1+\sin ^2 x\right) \cos x}{1+\sin ^4 x} d x \\ & \text{ Put } \sin x=t \\ & =\int \frac{1+t^2}{t^4+1} d t=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{\left(t-\frac{1}{t}\right)}{\sqrt{2}}+C \\ & x=\frac{\pi}{2}, t=1 \quad \therefore C=0 \\ & y\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(-\frac{1}{2}\right) \end{aligned}

Q48

If

\int \dfrac{\sin ^{\dfrac{3}{2}} x+\cos ^{\dfrac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x=A \sqrt{\cos \theta \tan x-\sin \theta}+B \sqrt{\cos \theta-\sin \theta \cot x}+C

, where

C

is the integration constant, then

A B

is equal to

A

2 \sec \theta

B

8 \operatorname{cosec}(2 \theta)

C

4 \operatorname{cosec}(2 \theta)

D

4 \sec \theta

Correct Answer

Option B

Solution

\begin{aligned} & \text{} \int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x \\ & I=\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x(\sin x \cos \theta-\cos x \sin \theta)}} d x \\ & =\int \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x \cos ^2 x \sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\cos ^{\frac{3}{2}} x}{\sin ^2 x \cos ^{\frac{3}{2}} x \sqrt{\cos \theta-\cot x \sin \theta}} d x= \\ & \int \frac{\sec ^2 x}{\sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\operatorname{cosec}^2 x}{\sqrt{\cos \theta-\cot x \sin \theta}} d x \\ & \end{aligned}

\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2

...... {Let} For

\mathrm{I}_1

, let

\tan \mathrm{x} \cos \theta-\sin \theta=\mathrm{t}^2

\sec ^2 x d x=\frac{2 t d t}{\cos \theta}

For

\mathrm{I}_2

, let

\cos \theta-\cot \mathrm{x} \sin \theta=\mathrm{z}^2

\operatorname{cosec}^2 x d x=\frac{2 z d z}{\sin \theta}

\begin{aligned} & I=I_1+I_2 \\ & =\int \frac{2 t d t}{\cos \theta t}+\int \frac{2 z d z}{\sin \theta z} \\ & =\frac{2 t}{\cos \theta}+\frac{2 z}{\sin \theta} \\ & =2 \sec \theta \sqrt{\tan x \cos \theta-\sin \theta}+2 \operatorname{cosec} \theta \sqrt{\cos \theta-\cot x \sin \theta} \\ & \text{ Comparing } \\ & \quad A B=8 \operatorname{cosec} 2 \theta \end{aligned}

Q49

Let

\int \dfrac{2-\tan x}{3+\tan x} \mathrm{~d} x=\dfrac{1}{2}\left(\alpha x+\log _e|\beta \sin x+\gamma \cos x|\right)+C

, where

C

is the constant of integration. Then

\alpha+\dfrac{\gamma}{\beta}

is equal to :

A 3

B 7

C 1

D 4

Correct Answer

Option D

Solution

I=\int \frac{2-\tan x}{3+\tan x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x

Put,

2 \cos x-\sin x=a(-3 \sin x+\cos x)+ b(3 \cos x+\sin x)

\begin{aligned} & a+3 b=2 \quad \text{.... (i)}\\ & -3 a+b=-1 \quad \text{.... (ii)} \end{aligned}

From equation (i) and (ii)

a=b=\frac{1}{2}

I=\frac{1}{2} \int \frac{-3 \sin x+\cos x}{3 \cos x+\sin x} d x+\frac{1}{2} \int \frac{3 \cos x+\sin x}{3 \cos x+\sin x} d x

I=\frac{1}{2} \ln |3 \cos x+\sin x|+\frac{1}{2} x= \frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|)

On comparing, we get

I=\frac{1}{2} \ln |3 \cos x+\sin x|+\frac{1}{2} x=\frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|)

On comparing, we get

\begin{aligned} & \Rightarrow \frac{1}{2}(x+\log (|3 \cos x+\sin x|))= \\ & \qquad \frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|) \end{aligned}

\begin{aligned} & \alpha=1, \beta=1, \gamma=3 \\ & \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4 \end{aligned}

Q50

Let

I(x)=\int \dfrac{6}{\sin ^2 x(1-\cot x)^2} d x

. If

I(0)=3

, then

I\left(\dfrac{\pi}{12}\right)

is equal to

A

\sqrt3

B

2\sqrt3

C

6\sqrt3

D

3\sqrt3

Correct Answer

Option D

Solution

\begin{aligned} & I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x \\ & I(x)=\int \frac{6}{(\sin x-\cos x)^2} d x \\ & =\int \frac{6 \sec ^2 x}{(\tan x-1)^2} d x \end{aligned}

\begin{aligned} & \text{ Let } \tan x=t \Rightarrow \sec ^2 x d x=d t \\ & =\int \frac{6 d t}{(t-1)^2} \\ & =-\frac{6}{(t-1)}+c \\ & =\frac{-6}{(\tan x-1)}+c \\ & I(x)=\frac{6}{1-\tan x}+c \\ & I(0)=3 \\ & \frac{6}{1-\tan 0}+c=3 \\ & c=-3 \end{aligned}

\begin{aligned} I(x) & =\frac{6}{1-\tan x}-3 \\ I\left(\frac{\pi}{12}\right) & =\frac{6}{1-\tan \left(\frac{\pi}{12}\right)}-3 \\ & =\frac{6}{1-(2-\sqrt{3})}-3 \\ & =\frac{6}{\sqrt{3}-1}-3 \\ & =\frac{6-3 \sqrt{3}+3}{\sqrt{3}-1} \\ & =\frac{9-3 \sqrt{3}}{\sqrt{3}-1} \\ & =\frac{3 \sqrt{3}(\sqrt{3}-1)}{\sqrt{3}-1} \\ & =3 \sqrt{3} \end{aligned}