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Cx + A

/ | C dx /

(We know that x^0 = 1)

/ = C| 1 . dx / / 0 = C| x . dx / / 0 = C| x . 1 dx /

(Here f(x) is x and 0 is its power, derivative of x is 1 which is also exist in front of x. So apply the rule.)

0 + 1 x = C ------- + A 0 + 1 1 x = C ----- + A 1 = Cx + A

(Here A is constant. We can also take C as constant but a constant C is already exit in question. So the new constant may or may not equal to C. If equal to C then we can take common C then answer may become C(x + 1).)

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(x^0.1/0.1, x^0.2/0.2, x^1.1/1.1, x^2.2/2.2)

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