Problem 3. $$ \int_{-3}^{3} \frac{x^{4}}{1+e^{-x}} \mathrm{~d} x $$ Problem 4. $$ \int_{0}^{1}(\sin (\log x)+\cos (\log x)) \mathrm{d} x $$ Problem 5. $$ \int_{-1}^{1} \frac{\cos (x)}{\arctan (x)} \mathrm{d} x $$

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**Problem 3**

We wish to evaluate
$$
I=\int_{-3}^{3}\frac{x^4}{1+e^{-x}}\,dx.
$$
Notice that the numerator $$x^4$$ is an even function. Although the denominator $$1+e^{-x}$$ is not even, we can use a symmetry trick. Define
$$
f(x)=\frac{x^4}{1+e^{-x}}
$$
and consider
$$
f(-x)=\frac{(-x)^4}{1+e^{x}}=\frac{x^4}{1+e^{x}}.
$$
Now, add the two:
$$
f(x)+f(-x)=\frac{x^4}{1+e^{-x}}+\frac{x^4}{1+e^{x}}.
$$
Factor out $$x^4$$:
$$
f(x)+f(-x)=x^4\left(\frac{1}{1+e^{-x}}+\frac{1}{1+e^{x}}\right).
$$
Observe that
$$
\frac{1}{1+e^{-x}} = \frac{e^x}{e^x+1},
$$
so that
$$
\frac{1}{1+e^{-x}}+\frac{1}{1+e^{x}}=\frac{e^x}{e^x+1}+\frac{1}{e^x+1}=\frac{e^x+1}{e^x+1}=1.
$$
Thus,
$$
f(x)+f(-x)=x^4.
$$
Since the interval is symmetric, we have
$$
\int_{-3}^{3}f(x)\,dx+\int_{-3}^{3}f(-x)\,dx=\int_{-3}^{3}x^4\,dx.
$$
But by a change of variable $$x\to -x$$ the two integrals on the left are equal, so
$$
2\int_{-3}^{3}f(x)\,dx=\int_{-3}^{3}x^4\,dx.
$$
Now, compute the even function integral:
$$
\int_{-3}^{3}x^4\,dx=2\int_{0}^{3}x^4\,dx =2\left[\frac{x^5}{5}\right]_{0}^{3}=2\cdot\frac{3^5}{5}=2\cdot\frac{243}{5}=\frac{486}{5}.
$$
Thus,
$$
2I=\frac{486}{5}\quad\Longrightarrow\quad I=\frac{243}{5}.
$$

---

**Problem 4**

We need to evaluate
$$
I=\int_{0}^{1}\bigl(\sin(\log x)+\cos(\log x)\bigr)\,dx.
$$
Apply the substitution
$$
u=\log x\quad\Longrightarrow\quad x=e^u,\quad dx=e^u\,du.
$$
When $$x=0$$, $$u\to-\infty$$; when $$x=1$$, $$u=0$$. The integral becomes
$$
I=\int_{-\infty}^{0}\bigl(\sin u+\cos u\bigr)e^u\,du.
$$
Separate the integral:
$$
I=\int_{-\infty}^{0}e^u\sin u\,du+\int_{-\infty}^{0}e^u\cos u\,du.
$$
Recall the standard integrals:
$$
\int e^{u}\sin u\,du=\frac{e^{u}}{2}(\sin u-\cos u)+C,
$$
$$
\int e^{u}\cos u\,du=\frac{e^{u}}{2}(\sin u+\cos u)+C.
$$
Adding these, we have
$$
\int e^{u}\sin u\,du+\int e^{u}\cos u\,du=\frac{e^{u}}{2}[(\sin u-\cos u)+(\sin u+\cos u)]=e^{u}\sin u+C.
$$
Thus,
$$
I=\left[e^{u}\sin u\right]_{u=-\infty}^{0}.
$$
At $$u=0$$:
$$
e^0\sin 0=1\cdot 0=0.
$$
At $$u\to -\infty$$:
$$
e^{u}\to 0,\quad \sin u \text{ is bounded, so } e^{u}\sin u\to 0.
$$
Hence,
$$
I=0-0=0.
$$

---

**Problem 5**

We wish to evaluate
$$
I=\int_{-1}^{1}\frac{\cos x}{\arctan x}\,dx.
$$
Examine the integrand. The numerator $$\cos x$$ is an even function since $$\cos(-x)=\cos x$$, and the denominator $$\arctan x$$ is an odd function because $$\arctan(-x)=-\arctan x$$. Therefore, the integrand satisfies
$$
\frac{\cos(-x)}{\arctan(-x)}=\frac{\cos x}{-\arctan x}=-\frac{\cos x}{\arctan x}.
$$
This shows that $$\frac{\cos x}{\arctan x}$$ is an odd function. An integral of an odd function over a symmetric interval about zero is zero provided the integral is interpreted as a Cauchy principal value:
$$
I=\int_{-1}^{1}\text{(odd function)}\,dx=0.
$$

---

$$
\boxed{
\begin{aligned}
\text{Problem 3:} &\quad \frac{243}{5},\[1mm]
\text{Problem 4:} &\quad 0,\[1mm]
\text{Problem 5:} &\quad 0\, (\text{as a principal value}).
\end{aligned}
}
$$
**Problem 3:** 243/5 **Problem 4:** 0 **Problem 5:** 0

Problem 3. \( \int_{-3}^{3} \frac{x^{4}}{1+e^{-x}} \mathrm{~d} x \) Problem 4. \( \int_{0}^{1}(\sin (\log x)+\cos (\log x)) \mathrm{d} x \) Problem 5. \( \int_{-1}^{1} \frac{\cos (x)}{\arctan (x)} \mathrm{d} x \)

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