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

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

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

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

= \int_{0}^{1} (1 + y + x)^{- 2} d x

Apply u-substitution

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

Apply the Power Rule

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

Compute the boundaries:

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

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

Simplify

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

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