Question

求下列不定积分

Answer 1

（3）原式=∫(e^x)dx/[e^x+e^(3x)]
=∫d(e^x)/[e^x+e^(3x)]
=∫{1/e^x-e^x/[1+e^(2x)]}d(e^x)
=ln(e^x)-(1/2)*ln[1+e^(2x)]+C，其中C是任意常数
（4）原式=∫2arctan(√x)/(1+x)d(√x)
=∫2arctan(√x)d[arctan(√x)]
=[arctan(√x)]^2+C，其中C是任意常数
（5）∫x[x-√(x^2-1)]/[x+√(x^2-1)][x-√(x^2-1)]dx
=∫[x^2-x√(x^2-1)]dx
=∫x^2dx-∫x√(x^2-1)dx
=(1/3)*x^3-(1/2)*∫√(x^2-1)d(x^2-1)
=(1/3)*x^3-(1/3)*(x^2-1)^(3/2)+C，其中C是任意常数
（6）令x=asint，则dx=acostdt
原式=∫acostdt/(asint+acost)
=∫costdt/(sint+cost)
令A=∫costdt/(sint+cost)，B=∫sintdt/(sint+cost)
则A+B=∫(cost+sint)dt/(sint+cost)
=∫dt
=t+C1，其中C1是任意常数
A-B=∫(cost-sint)dt/(sint+cost)
=∫d(sint+cost)/(sint+cost)
=ln|sint+cost|+C2，其中C2是任意常数
所以A=[(A+B)+(A-B)]/2
=t/2+(1/2)*ln|sint+cost|+C
原式=(1/2)*arcsin(x/a)+(1/2)*ln|x/a+√[1-(x/a)^2]|+C
=(1/2)*arcsin(x/a)+(1/2)*ln|x+√(a^2-x^2)|+C，其中C是任意常数