(a)
ex=1t−1ex=1t−1
x=−ln(t−1)
dx=−1t−1dt
∫1e4x+e3xdx
=∫1(1t−1)4+(1t−1)3⋅−1t−1dt
=−∫1(1t−1)3+(1t−1)2dt
=−∫(t−1)31+t−1dt
=−∫t3−3t2+3t−1tdt
=−13t3+32t2−3t+ln|t|+C
=−13(1ex+1)3+32(1ex+1)2−3(1ex+1)+ln|1ex+1|+C
(b)
Let u=ex . Then du=exdx .
∫1u5+u4du
=∫1e5x+e4x⋅exdx
=∫1e4x+e3xdx
=−13(1ex+1)3+32(1ex+1)2−3(1ex+1)+ln|1ex+1|+C ( by (a) )
=−13(1u+1)3+32(1u+1)2−3(1u+1)+ln|1u+1|+C
(c)
∫e15u5+4u4(u5+u4)2du
=∫e1(5u4+4u3)u(u5+u4)2du
=∫u=eu=1u(u5+u4)2d(u5+u4)
=∫u=eu=1u(u5+u4)−2d(u5+u4)
=−∫u=eu=1ud(u5+u4)−1
=−[u(u5+u4)−1]u=eu=1+∫u=eu=1(u5+u4)−1du
=−[(u4+u3)−1]u=eu=1+[−13(1u+1)3+32(1u+1)2−3(1u+1)+ln|1u+1|]u=eu=1
=−(e4+e3)−1−13(1e+1)3+32(1e+1)2−3(1e+1)+ln(1e+1)+196−ln2
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