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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

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

= - \frac{1}{x + 2} + \frac{1}{x + 1}

Simplify

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

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