Key Concepts and Formulas

• Integration by Parts: $\int u \, dv = uv - \int v \, du$
• Differentiation of Trigonometric Functions: $\frac{d}{dx}(\tan x) = \sec^2 x$, $\frac{d}{dx}(\sin x) = \cos x$, $\frac{d}{dx}(\cos x) = -\sin x$
• Logarithm Properties: $n \log a = \log a^n$

Step-by-Step Solution

Step 1: Rewrite the integral and prepare for integration by parts. We are given the integral $I(x)=\int \frac{x^{2}\left(x \sec ^{2} x+\tan x\right)}{(x \tan x+1)^{2}} d x$ The goal is to find a suitable $u$ and $dv$ for integration by parts. We want to choose $u$ and $dv$ such that the resulting integral is simpler.

Step 2: Apply integration by parts. Let's try to rewrite the integral as follows: $I(x) = \int x^2 \cdot \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} dx$ We will use integration by parts. Let $u = x^2$ and $dv = \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} dx$. Then, $du = 2x \, dx$. To find $v$, we need to integrate $dv$. Notice that the numerator of $dv$ is the derivative of $x \tan x + 1$, except for a constant factor. Let $w = x \tan x + 1$. Then $\frac{dw}{dx} = \tan x + x \sec^2 x$. So, we have $v = \int \frac{x \sec^2 x + \tan x}{(x \tan x + 1)^2} dx = \int \frac{dw}{w^2} = -\frac{1}{w} = -\frac{1}{x \tan x + 1}$ Now, applying integration by parts, we get: $I(x) = x^2 \left( -\frac{1}{x \tan x + 1} \right) - \int \left( -\frac{1}{x \tan x + 1} \right) (2x \, dx)$ $I(x) = -\frac{x^2}{x \tan x + 1} + \int \frac{2x}{x \tan x + 1} dx$

Step 3: Evaluate the remaining integral. Let $I_1 = \int \frac{2x}{x \tan x + 1} dx = 2 \int \frac{x}{x \tan x + 1} dx$. We can rewrite this as: $I_1 = 2 \int \frac{x \cos x}{x \sin x + \cos x} dx$ Now, let $t = x \sin x + \cos x$. Then $\frac{dt}{dx} = \sin x + x \cos x - \sin x = x \cos x$. So, $dt = x \cos x \, dx$. Therefore, $I_1 = 2 \int \frac{dt}{t} = 2 \ln |t| + C = 2 \ln |x \sin x + \cos x| + C$

Step 4: Substitute back into the expression for I(x). $I(x) = -\frac{x^2}{x \tan x + 1} + 2 \ln |x \sin x + \cos x| + C$

Step 5: Determine the constant of integration using the given initial condition. We are given that $I(0) = 0$. Substituting $x = 0$ into the expression for $I(x)$, we get: $I(0) = -\frac{0^2}{0 \cdot \tan 0 + 1} + 2 \ln |0 \cdot \sin 0 + \cos 0| + C = 0$ $0 = 0 + 2 \ln |0 + 1| + C = 2 \ln(1) + C = 0 + C$ Thus, $C = 0$.

Step 6: Write the final expression for I(x). $I(x) = -\frac{x^2}{x \tan x + 1} + 2 \ln |x \sin x + \cos x|$

Step 7: Evaluate I(π/4). We need to find $I(\frac{\pi}{4})$. Substituting $x = \frac{\pi}{4}$ into the expression for $I(x)$, we get: $I\left(\frac{\pi}{4}\right) = -\frac{\left(\frac{\pi}{4}\right)^2}{\frac{\pi}{4} \tan \left(\frac{\pi}{4}\right) + 1} + 2 \ln \left| \frac{\pi}{4} \sin \left(\frac{\pi}{4}\right) + \cos \left(\frac{\pi}{4}\right) \right|$ $I\left(\frac{\pi}{4}\right) = -\frac{\frac{\pi^2}{16}}{\frac{\pi}{4} (1) + 1} + 2 \ln \left| \frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right|$ $I\left(\frac{\pi}{4}\right) = -\frac{\frac{\pi^2}{16}}{\frac{\pi + 4}{4}} + 2 \ln \left| \frac{\pi + 4}{4\sqrt{2}} \right|$ $I\left(\frac{\pi}{4}\right) = -\frac{\pi^2}{16} \cdot \frac{4}{\pi + 4} + 2 \ln \left( \frac{\pi + 4}{4\sqrt{2}} \right)$ $I\left(\frac{\pi}{4}\right) = -\frac{\pi^2}{4(\pi + 4)} + \ln \left( \frac{(\pi + 4)^2}{(4\sqrt{2})^2} \right)$ $I\left(\frac{\pi}{4}\right) = -\frac{\pi^2}{4(\pi + 4)} + \ln \left( \frac{(\pi + 4)^2}{16 \cdot 2} \right)$ $I\left(\frac{\pi}{4}\right) = -\frac{\pi^2}{4(\pi + 4)} + \ln \left( \frac{(\pi + 4)^2}{32} \right)$

Common Mistakes & Tips

• Choosing u and dv: Careful selection of $u$ and $dv$ in integration by parts is crucial for simplifying the integral.
• Sign Errors: Pay close attention to sign errors when applying integration by parts and simplifying expressions.
• Logarithm Arguments: Remember that the argument of a logarithm must be positive. Use absolute value signs when necessary and remove them when the argument is guaranteed to be positive.

Summary

We found the integral $I(x)$ by applying integration by parts and then using a substitution to evaluate the remaining integral. We determined the constant of integration using the given initial condition $I(0) = 0$. Finally, we evaluated $I(\frac{\pi}{4})$ to obtain the desired result.

The final answer is \boxed{\log _{e} \frac{(\pi+4)^{2}}{32}-\frac{\pi^{2}}{4(\pi+4)}}, which corresponds to option (A).