∫1 sinx 1 cos∧2x

答案见附图

∫(x+1)/(x^2-2x+5)dx=1/2*∫(2x-2)/(x^2-2x+5)dx+∫2/(x^2-2x+5)dx=1/2*∫[1/(x^2-2x+5)]d(x^2-2x+5)+2∫1/[(x-

原式=-ln(1+x)/x+∫dx/[x(1+x)](应用分部积分法)=-ln(1+x)/x+∫[1/x-1/(1+x)]dx=-ln(1+x)/x+ln│x│-ln(1+x)+C(C是任意常数).

你是问为什么∫du/(u²+a²)²=(1/2a²)[u/(u²+a²)+∫du/(u²+a²)],对吗?如果是这么个问

∫（x^4/(x^2+1)）dx=∫（(x^4-1+1)/(x^2+1)）dx=∫（(x^4-1)/(x^2+1)）dx+∫（1/(x^2+1)）dx=∫（(x^2-1)*(x^2+1)/(x^2+1

再问：详细点再问：这变的好突然再答：哦，看来是这不懂啊，把分母拆开，二次项那个分子设为AX+B，一次项那个设为C，两个分式相加分子等于x，就可以把ABC解出来了再问：好深奥的样子。。。我问问数学课代表

亲爱的楼主：∫(1－1/x^2)e^(x＋1/x)dx其中因为(x＋1/x)'=1-1/x^2则d(x＋1/x)=(1-1/x^2)dx原式=∫e^(x＋1/x)d(x＋1/x)=e^(x＋1/x)+

左边=∫x√(1-x^2)dx-∫x^2dx=-1/2∫√(1-x^2)d(-x^2)-x^3/3=-1/2*2/3*(1-x^2)^(3/2)-x^3/3+C=-1/3*(1-x^2)^(3/2)-

当x=0时,f(x)不连续,故f(x)的原函数分成两部分：x>0,∫f(x)dx=∫x㏑(1+x^2)dx=(1/2)∫㏑(1+x^2)d(x^2)=(1/2)ln|ln(1+x^2)|+C1x

你好∫(1+2x)/x(1+x)*dx=∫(1+x+x)/x(1+x)*dx=∫[1/(1+x)+1/x]*dx=ln(x+1)+lnx+C很高兴为您解答,祝你学习进步!有不明白的可以追问!如果有其他

∫（3x+1/x∧2)dx=-∫（3x+1)d(1/x)=-(3x+1)/x+∫（1/x)d(3x+1)=-(3x+1)/x+1/3∫（1/x)dx=-(3x+1)/x+1/3ln|x|+c回答完毕!

∫x/[(x^2+1)(x^2+4)]dx=1/3∫x[1/(x^2+1)-1/(x^2+4)]dx=1/3[∫x/(x^2+1)dx-∫x/(x^2+4)dx]=1/3[1/2∫1/[(x^2+1)

∫f'(x)dx/1+f^2(x)=∫df(x)/[1+f^2(x)]=arctanf(x)+c=arctan(e^x/x)+c

∵(x^4-4x^2+5x-15)/[(x^2+1)(x-2)]=[(x^4+x²-5x²-5)+(5x-10)]/[(x²+1)(x-2)]=[x²(x&su

(3x-2)（3x+2)-x(5x+1)>(2x+1)²9x²-4-5x²-x>4x²+4x+14x+1

答：f(x)=x^3-x(-2x+x^2-1)=x^3+2x^2-x^3+x=2x^2+x=2(x+1/4)^2-1/8当且仅当x=-1/4时,f(x)取得最小值-1/8

令x=tanu,则dx=sec²tdt∫1/[x√(1+x²)]dx=∫1/[tanu·√(1+tan²x)]·sec²tdt=∫cscudu=-ln|cscu

[(x^3-2x^2+x+1)/(x^4+5x^2+4)]=1/(x^2+1)+(x-3)/(x^2+4).原式=∫1/(x^2+1)dx+∫(x-3)/(x^2+4)dx=arctanx+(1/2)