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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).)