求不定积分 ∫ln(x^2+a^2)dx

用分部积分法：∫udv=uv-∫vdu
∫ln(x²+a²)dx
=xln(x²+a²)-∫xd(ln(x²+a²))
=xln(x²+a²)-∫x*2x/(x²+a²) dx
=xln(x²+a²)-2∫(x²+a²-a²)/(x²+a²)dx
=xln(x²+a²)-2∫[1-a²/(x²+a²)]dx
=xln(x²+a²)-2x+2a²∫1/(x²+a²)dx
=xln(x²+a²)-2x+2a²*arctan(x/a)/a+C
=xln(x²+a²)-2x+2aarctan(x/a)+C

唉 不会呀

分部积分=∫(x)'ln(x^2+a^2)=xln(*)-∫xdln(*)=xln(*)-2x+2a*arctan(x/a)+C…剩下的就知道了吧？