What is the integral of 1/((4x-5)^2) ?

The integral of 1/((4x-5)^2) is -1/(4(4x-5))+C

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integral of 1/((4x-5)^2)

\int \frac{1}{( 4 x - 5 ) ^{2}} d x

- \frac{1}{4 ( 4 x - 5 )} + C

Solution steps

\int \frac{1}{( 4 x - 5 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{4 u ^{2}} d u

= \int \frac{1}{4 u ^{2}} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{4} (- \frac{1}{u})

Substitute back

u = 4 x - 5

= \frac{1}{4} (- \frac{1}{4 x - 5})

Simplify

\frac{1}{4} (- \frac{1}{4 x - 5}) : - \frac{1}{4 ( 4 x - 5 )}

= - \frac{1}{4 ( 4 x - 5 )}

Add a constant to the solution

= - \frac{1}{4 ( 4 x - 5 )} + C

n = 1 \sum \infty (1 + \frac{5}{n})^{n}

f (x) = 3 x sin^{2} (x)

8 x + \frac{12800}{x}

\frac{d y}{d x} = 6 x

\int - \frac{1}{n} cos (n t) d t

What is the integral of 1/((4x-5)^2) ?
The integral of 1/((4x-5)^2) is -1/(4(4x-5))+C