What is the integral of (4x+6)^{-2} ?

The integral of (4x+6)^{-2} is -1/(4(4x+6))+C

Popular >

integral of (4x+6)^{-2}

\int (4 x + 6)^{- 2} d x

- \frac{1}{4 ( 4 x + 6 )} + C

Solution steps

\int (4 x + 6)^{- 2} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int \frac{1}{( 4 x + 6 ) ^{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 + 6

= \frac{1}{4} (- \frac{1}{4 x + 6})

Simplify

\frac{1}{4} (- \frac{1}{4 x + 6}) : - \frac{1}{4 ( 4 x + 6 )}

= - \frac{1}{4 ( 4 x + 6 )}

Add a constant to the solution

= - \frac{1}{4 ( 4 x + 6 )} + C

\int \frac{cos ( \frac{π}{x ^{23}} )}{x ^{24}} d x

\int (3 x^{4} - 5 x^{3} - x - 6) d x

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

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

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

What is the integral of (4x+6)^{-2} ?
The integral of (4x+6)^{-2} is -1/(4(4x+6))+C