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rotaxmax
@rotaxmax
September 2021
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∫㏑5x dx
∫(7x⁶-2\x⁵+²√2)dx
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Vasily1975
Verified answer
1) ∫ln(5*x)*dx, применяем метод интегрирования "по частям".
u(x)=ln(5*x), dv(x)=dx, du=dx/x, v=∫dx=x, ∫(ln(5*x)*dx=u*v-∫v*du=x*ln(5*x)-∫x*dx/x=x*ln(5*x)-∫dx=x*ln(5*x)-x+C. Ответ: x*ln(5*x)-x+C.
2) ∫(7*x⁶-2/x⁵+√2)*dx=7*∫x⁶*dx-2*∫dx/x⁵+√2*∫dx=7*x⁷/7-2*x^(-4)/(-4)+√2*x+C=x⁷+1/2*x⁴+√2*x+C. Ответ:
x⁷+1/2*x⁴+√2*x+C.
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Answers & Comments
Verified answer
1) ∫ln(5*x)*dx, применяем метод интегрирования "по частям".u(x)=ln(5*x), dv(x)=dx, du=dx/x, v=∫dx=x, ∫(ln(5*x)*dx=u*v-∫v*du=x*ln(5*x)-∫x*dx/x=x*ln(5*x)-∫dx=x*ln(5*x)-x+C. Ответ: x*ln(5*x)-x+C.
2) ∫(7*x⁶-2/x⁵+√2)*dx=7*∫x⁶*dx-2*∫dx/x⁵+√2*∫dx=7*x⁷/7-2*x^(-4)/(-4)+√2*x+C=x⁷+1/2*x⁴+√2*x+C. Ответ: x⁷+1/2*x⁴+√2*x+C.