What is the integral of 1/((2x+6)^2) ?

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

integral of 1/((2x+6)^2)

\int \frac{1}{( 2 x + 6 ) ^{2}} d x

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

Solution steps

\int \frac{1}{( 2 x + 6 ) ^{2}} d x

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = 2 x + 6

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

Simplify

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

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

Add a constant to the solution

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

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