如图:
原式=(1/2)∫ln(1+p²)d(p²) =(1/2)∫ln(1+p²)d(1+p²) =(1/2)(1+p²)ln(1+p²)-(1/2)∫d(1+p²) =(1/2)(1+p²)ln(1+p²)-(1/2)p²+C