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More about usThis integration trick can be applied when f(x) has a power n (here n ≠ 1) And derivate of f(x) is also persent of front of [f(x)]^n. Keep in mind always that n should not be equal to -1

In such case just take the f(x), take its power with plus 1 as power and divide by new power.

Let us we apply this integration rule on following examples.

/ 2 | x dx /

/ 2 = | x . 1 dx /

(Here f(x) is x and 2 is its power, derivative of x is 1 which is also exist in front of x. So apply the above rule. Just take f(x), take its power with plus 1 as power and divide by new power.)

2 + 1 x = ------- + C 2 + 1 3 x = ----- + C 3

/ | 1 dx /

/ 0 = | x . dx / / 0 = | 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 above rule. Just take f(x), take its power with plus 1 as power and divide by new power.)

0 + 1 x = ------- + C 0 + 1 1 x = ----- + C 1 = x + C

/ | ln(x) | ----- dx | x /

/ | 1 = | ln(x) . --- dx | x /

(Here f(x) is ln(x) and 1 is its power, derivative of ln(x) is 1/x which is also exist in front of ln(x).So apply the above integration rule. Just take f(x), take its power with plus 1 as power and divide by new power.)

1 + 1 (ln(x)) = ------- + C 1 + 1 2 (ln x) = ----- + C 2

/ | 1/2 | (2x + 3) dx /

/ 1 | 1/2 = ---| (2x + 3) . 2 dx 2 | /

(Here f(x) is 2x + 3 and 1/2 is its power, its derivative was not exit so we try to make its derivative by multiply and divided by 2. Now derivative of 2x + 3 is 2 which is in front of 2x + 3. So apply the above rule. Just take f(x), take its power with plus 1 as power and divide by new power.)

1 --- + 1 2 1 (2x + 3) = --- -------- + C 2 1 --- + 1 2 3 --- 2 1 (2x + 3) = --- -------- + C 2 3 --- 2 3/2 (2x + 3) = -------- 3

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

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