<s>[INST] Integrate $\int_{0}^{\pi/2}{\frac{x\cos\left(x\right)-\sin\left(x\right)}{\sin\left(x\right)+x^2}}dx$ 
$$\int_{0}^{\pi/2}{\dfrac{x\cos\left(x\right)-\sin\left(x\right)}{\sin\left(x\right)+x^2}}dx$$

I am unable to exploit the properties of definite integral, neither it seems that indefinite integration is possible.
 [/INST] I think that power 2 of $\sin x$ is missed and solve the other one as below. At the first sight, the numerator of the integrand seems to be an antiderivative of $x\sin x$. However, the negative sign in between reminds me that it should be an antiderivative of $\dfrac{\sin x}{x}$. So I decided to divide both numerator and denominator of the integrand by $x^2$ and it works wonderfully.
$\begin{aligned} \int_{0}^{\frac{\pi}{2}} \frac{x \cos x-\sin x}{\sin ^{2} x+x^{2}} d x =& \int_{0}^{\frac{\pi}{2}} \frac{\frac{x \cos x-\sin x}{x^{2}}}{\frac{\sin ^{2} x}{x^{2}}+1}  d x \\=& \int_{0}^{\frac{\pi}{2}} \frac{d\left(\frac{\sin x}{x}\right)}{\left(\frac{\sin x}{x}\right)^{2}+1} \\=&\left[\tan ^{-1}\left(\frac{\sin x}{x}\right)\right]_{0}^{\frac{\pi}{2}} \\=& \tan ^{-1}\left(\frac{2}{\pi}\right)-\frac{\pi}{4} \quad\left(\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right) \end{aligned}$
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