1. Interpret the integral

The given integral is

$$I=\int_{-\pi/2}^{\pi/2}\left(x^2+\ln\frac{\pi+x}{\pi-x}\right)\cos x\,dx.$$

We split it as

$$I=I_1+I_2,$$

where

$$I_1=\int_{-\pi/2}^{\pi/2} x^2\cos x\,dx, \qquad I_2=\int_{-\pi/2}^{\pi/2} \ln\frac{\pi+x}{\pi-x}\cos x\,dx.$$

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2. Evaluate $I_1$

Notice that $x^2$ is even and $\cos x$ is even, so $x^2\cos x$ is even. Hence

$$I_1=2\int_0^{\pi/2} x^2\cos x\,dx.$$

Now integrate by parts:

Let

$$u=x^2,\quad dw=\cos x\,dx.$$

Then

$$du=2x\,dx,\quad w=\sin x.$$

So

$$\int x^2\cos x\,dx=x^2\sin x-\int 2x\sin x\,dx.$$

Again integrate by parts on $\int 2x\sin x\,dx$:

$$\int x\sin x\,dx=-x\cos x+\sin x.$$

Thus

$$\int x^2\cos x\,dx=x^2\sin x+2x\cos x-2\sin x.$$

Now evaluate from $0$ to $\pi/2$:

$$\int_0^{\pi/2} x^2\cos x\,dx =\left[x^2\sin x+2x\cos x-2\sin x\right]_0^{\pi/2}.$$

At $x=\pi/2$,

$$\sin\frac\pi2=1,\quad \cos\frac\pi2=0,$$

so value is

$$\frac{\pi^2}{4}-2.$$

At $x=0$, value is $0$.

Therefore

$$\int_0^{\pi/2} x^2\cos x\,dx=\frac{\pi^2}{4}-2,$$

and hence

$$I_1=2\left(\frac{\pi^2}{4}-2\right)=\frac{\pi^2}{2}-4.$$

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3. Evaluate $I_2$ using symmetry

Consider

$$f(x)=\ln\frac{\pi+x}{\pi-x}.$$

Then

$$f(-x)=\ln\frac{\pi-x}{\pi+x}=-\ln\frac{\pi+x}{\pi-x}=-f(x).$$

So $f(x)$ is odd.

Also, $\cos x$ is even. Therefore,

$$f(x)\cos x$$

is odd.

The integral of an odd function over $[-a,a]$ is $0$. Hence

$$I_2=\int_{-\pi/2}^{\pi/2} \ln\frac{\pi+x}{\pi-x}\cos x\,dx=0.$$

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4. Combine the results

Thus

$$I=I_1+I_2=\left(\frac{\pi^2}{2}-4\right)+0=\frac{\pi^2}{2}-4.$$

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5. Check options

The value is

$$\boxed{\frac{\pi^2}{2}-4}.$$

This matches Option B.

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6. Compare with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.