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

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

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

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

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{2}{3}

= \frac{2}{3}

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