Integral of [x+3/9-x²]dx?

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Answer 1 · 2018-04-21T04:02:50

#1/2x^2+3/9x-1/3x^3+C#

To take the integral of this expression, you should use the reverse power rule.

#inta^n=(a^(n+1))/(n+1)#

In this case, you have three terms: #x#, #3/9#, and #-x^2#

The integral of #x# by itself is

#intx^1=(x^(1+1))/(1+1)=x^2/2#

#3/9# which should be seen as #3/9*x^0# has the integral

#int3/9*x^0=3/9x^(0+1)/(0+1)=3/9x#

Finally, the integral of #-x^2# is

#int(-x^2)=-x^(2+1)/(2+1)=-x^3/3#

Combining all these terms and an unknown constant (C) you get the final answer:

#int(x+3/9-x^2)dx=1/2x^2+3/9x-1/3x^3+C#