What is the integral from 3 to 4 of 1/((x+y)^2) ?

The integral from 3 to 4 of 1/((x+y)^2) is 1/((y+4)(y+3))

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integral from 3 to 4 of 1/((x+y)^2)

\int_{3}^{4} \frac{1}{( x + y ) ^{2}} d x

\frac{1}{( y + 4 ) ( y + 3 )}

Solution steps

\int_{3}^{4} \frac{1}{( x + y ) ^{2}} d x

Apply u-substitution

: \int_{y + 3}^{y + 4} \frac{1}{u ^{2}} d u

= \int_{y + 3}^{y + 4} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{y + 3}^{y + 4}

= [- \frac{1}{u}]_{y + 3}^{y + 4}

Compute the boundaries:

- \frac{1}{y + 4} + \frac{1}{y + 3}

= - \frac{1}{y + 4} + \frac{1}{y + 3}

Simplify

= \frac{1}{( y + 4 ) ( y + 3 )}

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What is the integral from 3 to 4 of 1/((x+y)^2) ?
The integral from 3 to 4 of 1/((x+y)^2) is 1/((y+4)(y+3))