1. -(3²⁻¹¹ˣ)/(11㏑3)+C
2.
Пошаговое объяснение:ᵃ
1.∫3²⁻¹¹ˣdx=∫3ᵃda=-(3ᵃ)/(11㏑3)+C=-(3²⁻¹¹ˣ)/(11㏑3)+C
a=2-11x
da=-11dx
2.∫dx=∫(3a²(a³+1))/(a+1)da=3∫(a²(a³+1))/(a+1)da=3∫a²(a²-a+1)da=3∫(a⁴-a³+a²)da=3∫a⁴da-3∫a³da+3∫a²da==
a=∛x
da=dx
1. -(3²⁻¹¹ˣ)/(11㏑3)+C
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Пошаговое объяснение:ᵃ
1.∫3²⁻¹¹ˣdx=
∫3ᵃda=-(3ᵃ)/(11㏑3)+C=-(3²⁻¹¹ˣ)/(11㏑3)+C
a=2-11x
da=-11dx
2.∫
dx=∫(3a²(a³+1))/(a+1)da=3∫(a²(a³+1))/(a+1)da=3∫a²(a²-a+1)da=3∫(a⁴-a³+a²)da=3∫a⁴da-3∫a³da+3∫a²da=
=
a=∛x
da=
dx