What is the integral of 1/((2x+3y)^2) ?

The integral of 1/((2x+3y)^2) is -1/(2(2x+3y))+C

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integral of 1/((2x+3y)^2)

\int \frac{1}{( 2 x + 3 y ) ^{2}} d x

- \frac{1}{2 ( 2 x + 3 y )} + C

Solution steps

\int \frac{1}{( 2 x + 3 y ) ^{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 + 3 y

= \frac{1}{2} (- \frac{1}{2 x + 3 y})

Simplify

\frac{1}{2} (- \frac{1}{2 x + 3 y}) : - \frac{1}{2 ( 2 x + 3 y )}

= - \frac{1}{2 ( 2 x + 3 y )}

Add a constant to the solution

= - \frac{1}{2 ( 2 x + 3 y )} + C

\int sin^{2} (3 x co) s^{2} 3 x d x

\frac{d y}{d x} = e^{y + x}

\frac{d}{d v} (u sin (v))

\frac{\partial}{\partial y} (ln (5 + x^{2} y^{2}))

f (x) = 4 x - x^{2}, at x = 2

What is the integral of 1/((2x+3y)^2) ?
The integral of 1/((2x+3y)^2) is -1/(2(2x+3y))+C