求不定积分∫xln(1+x^2)dx

求不定积分∫xln(1+x^2)dx
求不定积分∫xln(1+x^2)dx

求不定积分∫xln(1+x^2)dx
∫xln(1+x^2)dx
=1/2∫ln(1+x^2)dx^2
=1/2∫ln(1+x^2)d(1+x^2)
=1/2(1+x^2)ln(1+x^2)-1/2∫(1+x^2)dln(1+x^2)
=1/2(1+x^2)ln(1+x^2)-1/2∫(1+x^2)*1/(1+x^2)d(1+x^2)
=1/2(1+x^2)ln(1+x^2)-1/2∫dx^2
=1/2(1+x^2)ln(1+x^2)-1/2x^2+C

令u=x^2,则du=2xdx,∫xln(1+x^2)dx=（1/2)∫ln(1+u)du,然后用分步积分就行了

∫xln(1+x^2)dx=∫1/2*ln(1+x^2)d(x^2+1)=1/2*(x^2+1)*(ln(x^2+1)-1)