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∫ln 1 x 2 dx

原式=xln(1+x^2)-∫xd[ln(1+x^2)]=xln(1+x^2)-∫2x^2/(1+x^2)dx=xln(1+x^2)-2∫[1-1/(1+x^2)dx=xln(1+x^2)-2x+2acrtgx+C

分部积分:∫ln(1+x^2)dx=xln(1+x^2)-∫2x^2/(1+x^2)dx=xln(1+x^2)-[∫2dx-∫2/(1+x^2)dx]=xln(1+x^2)-2x+2arctanx+c

【俊狼猎英】团队为您解答直接分部积分原积分=xln(1+x^2)-∫2x^2/(1+x^2)dx=xln(1+x^2)-2x+2∫dx/(1+x^2)=2acrtanx+xln(1+x^2)-2x+C

分步积分 ∫ln(1+x^2)dx =x*ln(1+x^2)-∫2x^2/(1+x^2)dx 对后面的进行分离 =x*ln(1+x^2)-∫2dx+∫2/(1+x^2)dx 直接积分 =x*ln(1+x^2)-2x+2arctanx+C du = 1/1+x^2dx ???错了吧 应该是2x/1+x^2dx 是复合函数,你还得对x^2求导啊.

=∫ln(1-x)dx+∫ln(1+x)d(1+x) =-∫ln(1-x)d(1-x)+[(1+x)ln(1+x)-x](0,1) =[-(1-x)ln(1-x)-x](0,1)+ln2-1 =-1-lim(x趋于1)(1-x)ln(1-x)+ln2-1 =ln2-2

答案肯定不对,ln后面不可能是x,而是x-1原式=-∫ln(x-1)d(1/x)=-ln(x-1)/x+∫(1/x)dln(x-1)=-ln(x-1)/x+∫1/x(x-1)dx=-ln(x-1)/x+∫[1/x-1/(x-1)]dx=-ln(x-1)/x+lnx-ln(x-1)+C

先化简被积函数∫ln(1/(1-x^2))dx=∫[-ln(1-x) - ln(1+x)] dx对两部分分别用分部积分=-xln(1-x)-xln(1+x) + ∫x/(x-1) dx + ∫x/(x+1) dx=-xln(1-x^2) + ∫[1 + 1/(x-1)] dx + ∫[1 - 1/(x+1)]dx=-xln(1-x^2) + 2x + ln|(x-1)/(x+1)| + C=-xln(1-x^2) + 2x + ln[(1-x)/(1+x)] + C

用分部积分法: ∫(0,1)ln(1+x^2)dx =x ln(1+x^2)|(0,1)- ∫(0,1)xdln(1+x^2) =ln2 - ∫(0,1)2x^2/(1+x^2)dx =ln2 - 2∫(0,1)[1 - 1/(1+x^2)]dx =ln2 - 2(x - arctanx))|(0,1) =ln2 - 2 + π/2

∫ln(1-x^2)dx=xln(1-x^2)-∫xdln(1-x^2)=xln(1-x^2)-∫x/(1-x^2)*(-2x)dx=xln(1-x^2)-2∫(-x^2)/(1-x^2)dx=xln(1-x^2)-2∫(1-x^2-1)/(1-x^2)dx=xln(1-x^2)-2∫dx+2∫1/(1-x^2)dx=xln(1-x^2)-2∫dx+∫[1/(1-x)+1/(1+x)]dx=xln(1-x^2)-2x+ln(1+x)-ln(1-x)+C积分限为0≤x≤1,则∫

分部积分 ∫ln(1+x^2)dx,= x ln(1+x^2) - ∫[x 1/(1+x^2) 2x] dx= x ln(1+x^2) - 2 [∫[1 - 1/(1+x^2)] dx]= x ln(1+x^2) - 2 x + 2 arctan(x) + C

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