integration-indefinite
Evaluate
¡Ò dx/3(2x-1)^2
(1/3) integral (2x-1)^-2 dx
let z = 2x-1
then dz = 2 dx
(1/3)(1/2) integral z^-2 dz
(1/6) (-1)z^-1
(-1/6) (2x-1)^-1
-1/[ 6(2x-1) ] + C
To evaluate the integral ∫ (1/3)(2x - 1)^2 dx, we can use the power rule of integration. The power rule states that ∫ x^n dx = (1/(n+1))x^(n+1) + C, where C is the constant of integration.
In this case, we have a quadratic term (2x - 1)^2, so we can expand the expression and then apply the power rule.
Step 1: Expand the expression
(2x - 1)^2 = (2x - 1)(2x - 1) = 4x^2 - 2x - 2x + 1 = 4x^2 - 4x + 1
Step 2: Apply the power rule
∫ (1/3)(2x - 1)^2 dx = (1/3)∫ (4x^2 - 4x + 1) dx
= (1/3) * [(4/3)x^3 - 2x^2 + x] + C
= (4/9)x^3 - (2/3)x^2 + (1/3)x + C
So, the value of the integral is (4/9)x^3 - (2/3)x^2 + (1/3)x + C, where C is the constant of integration.