Question

求不定积分 ∫ln(x^2+a^2)dx

Answer 1

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