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

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

Popular >

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

= \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}{u}

= \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

\int (x + 1) 2 x - 1 d x

\frac{d x}{d t} = x^{2}, x (0) = 1

\int - 6 x^{- 7} d x

\frac{d y}{d x} + (\frac{1 - y ^{2}}{1 - x ^{2}})^{\frac{1}{2}} = 0

\frac{d y}{d x} = 9 y

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