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

The integral from 1 to 2 of 1/((2x+3y)^2) is 1/(2(x+3)(2x+3))

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

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

\frac{1}{2 ( x + 3 ) ( 2 x + 3 )}

Solution steps

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

Apply u-substitution

: \int_{2 x + 3}^{2 x + 6} \frac{1}{3 u ^{2}} d u

= \int_{2 x + 3}^{2 x + 6} \frac{1}{3 u ^{2}} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{2 x + 3}^{2 x + 6} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{2 x + 3}^{2 x + 6}

= \frac{1}{3} [- \frac{1}{u}]_{2 x + 3}^{2 x + 6}

Compute the boundaries:

- \frac{1}{2 x + 6} + \frac{1}{2 x + 3}

= \frac{1}{3} (- \frac{1}{2 x + 6} + \frac{1}{2 x + 3})

Simplify

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

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