What is the integral of 1/((8x-1)^2) ?

The integral of 1/((8x-1)^2) is -1/(8(8x-1))+C

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integral of 1/((8x-1)^2)

\int \frac{1}{( 8 x - 1 ) ^{2}} d x

- \frac{1}{8 ( 8 x - 1 )} + C

Solution steps

\int \frac{1}{( 8 x - 1 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{8 u ^{2}} d u

= \int \frac{1}{8 u ^{2}} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{8} (- \frac{1}{u})

Substitute back

u = 8 x - 1

= \frac{1}{8} (- \frac{1}{8 x - 1})

Simplify

\frac{1}{8} (- \frac{1}{8 x - 1}) : - \frac{1}{8 ( 8 x - 1 )}

= - \frac{1}{8 ( 8 x - 1 )}

Add a constant to the solution

= - \frac{1}{8 ( 8 x - 1 )} + C

n = 1 \sum \infty (- 1)^{n} (\frac{1}{n})

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What is the integral of 1/((8x-1)^2) ?
The integral of 1/((8x-1)^2) is -1/(8(8x-1))+C