G(x) = ∫f(t)dt t =0→u(x)
则 G'(x) = f(x) * u'(x)
= ln(x²+5) * [e^(2x)]'
= ln(x²+5) * [2*e^(2x)]
= 2*ln(x²+5)*e^(2x)
设 u = e^x。则 du = e^x * dx
那么，上面的不定积分
=∫(4-e^x)*(e^x * dx)/(e^x + 3)
=∫(4-u)*du/(u+3)
=∫[7-(u+3)]*du/(u+3)
=∫7*du/(u+3) - ∫(u+3)*du/(u+3)
=7*∫du/(u+3) - ∫du
=7*∫d(u+3)/(u+3) - u
=7*ln(u+3) - u + C
=7*ln(e^x + 3) - e^x + C
温馨提示：答案为网友推荐，仅供参考
相似回答