What is the integral of 1/(3(2x-1)^2) ?

The integral of 1/(3(2x-1)^2) is -1/(6(2x-1))+C

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integral of 1/(3(2x-1)^2)

\int \frac{1}{3 ( 2 x - 1 ) ^{2}} d x

- \frac{1}{6 ( 2 x - 1 )} + C

Solution steps

\int \frac{1}{3 ( 2 x - 1 ) ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1}{( 2 x - 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{2 u ^{2}} d u

= \frac{1}{3} \cdot \int \frac{1}{2 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{3} \cdot \frac{1}{2} (- \frac{1}{u})

Substitute back

u = 2 x - 1

= \frac{1}{3} \cdot \frac{1}{2} (- \frac{1}{2 x - 1})

Simplify

\frac{1}{3} \cdot \frac{1}{2} (- \frac{1}{2 x - 1}) : - \frac{1}{6 ( 2 x - 1 )}

= - \frac{1}{6 ( 2 x - 1 )}

Add a constant to the solution

= - \frac{1}{6 ( 2 x - 1 )} + C

\int_{0}^{2} e^{x} - e^{- x} d x

g (x) = \frac{4 x ^{2} - x + 2}{3 - 4 x}

y = 3 x^{2} + 2 xx a

x \to 0 lim (x^{6} (cos (\frac{9}{x})))

What is the integral of 1/(3(2x-1)^2) ?
The integral of 1/(3(2x-1)^2) is -1/(6(2x-1))+C