What is the integral from 0 to 1 of 1/((2x+3y)^2) ?

The integral from 0 to 1 of 1/((2x+3y)^2) is 1/(3y(3y+2))

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integral from 0 to 1 of 1/((2x+3y)^2)

\int_{0}^{1} \frac{1}{( 2 x + 3 y ) ^{2}} d x

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

Solution steps

\int_{0}^{1} \frac{1}{( 2 x + 3 y ) ^{2}} d x

Apply u-substitution

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

= \int_{3 y}^{2 + 3 y} \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_{3 y}^{2 + 3 y} \frac{1}{u ^{2}} d u

Apply the Power Rule

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

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

Compute the boundaries:

- \frac{1}{2 + 3 y} + \frac{1}{3 y}

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

Simplify

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

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What is the integral from 0 to 1 of 1/((2x+3y)^2) ?
The integral from 0 to 1 of 1/((2x+3y)^2) is 1/(3y(3y+2))