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1.求不定积分∫(x^2)dx/(x-1),∫xsin(x^2+1)dx2.已知∫f(x)dx=xe^x+才,则f(x)=

题目详情
1.求不定积分∫(x^2)dx/(x-1),∫xsin(x^2+1)dx 2.已知∫f(x)dx=xe^x+才,则f(x)=
▼优质解答
答案和解析
∫(x^2)dx/(x-1)
=∫((x^2)-1+1)dx/(x-1)
∫(x+1+1/(x-1))dx =1/2*x^2 +x +ln(x-1)+C
∫xsin(x^2+1)dx
=1/2∫sin(x^2+1)d(x^2+1)
=-1/2cos(x^2+1)
∫f(x)dx=xe^x
f(x) =(xe^x)'=(x+1)e^x