Choose parts, repeat it, or solve the returning integral

Choose $u=\ln x$, $dv/dx=2x+3$, $du/dx=1/x$, $v=x^2+3x$. Then $\int(2x+3)\ln x\,dx=(x^2+3x)\ln x-\int(x+3)\,dx=(x^2+3x)\ln x-x^2/2-3x+C$, for $x>0$.

Choose $u=x$, $dv/dx=\sin x$, $du/dx=1$, $v=-\cos x$. The result is $-x\cos x-\int(-\cos x)\,dx=-x\cos x+\sin x+C$.

Write $\ln x=(\ln x)(1)$, then choose $u=\ln x$, $dv/dx=1$, $du/dx=1/x$, $v=x$. Hence $\int\ln x\,dx=x\ln x-\int x(1/x)\,dx=x\ln x-x+C$, for $x>0$. This is the source important remark about introducing a factor 1.

First $u=x^2$, $dv/dx=\cos x$, $du/dx=2x$, $v=\sin x$, giving $x^2\sin x-2\int x\sin x\,dx$. For the latter integral use $u_1=x$, $dv_1/dx=\sin x$, $du_1/dx=1$, $v_1=-\cos x$. Therefore the answer is $x^2\sin x+2x\cos x-2\int\cos x\,dx=x^2\sin x+2x\cos x-2\sin x+C$.

Let $I=\int e^{2x}\sin x\,dx$. Choose $u=\sin x$, $dv/dx=e^{2x}$, so $du/dx=\cos x$, $v=e^{2x}/2$. Then $I=e^{2x}\sin x/2-\frac12\int e^{2x}\cos x\,dx$. Apply parts again with $u_1=\cos x$, $du_1/dx=-\sin x$, $dv_1/dx=e^{2x}$, $v_1=e^{2x}/2$. Thus $I=e^{2x}\sin x/2-e^{2x}\cos x/4-I/4$. Collect it on the left: $5I/4=e^{2x}\sin x/2-e^{2x}\cos x/4+C_1$, giving $I=\frac25e^{2x}\sin x-\frac15e^{2x}\cos x+C$.