What is the integral from 0 to 9 of 1/((1+x)^2) ?

The integral from 0 to 9 of 1/((1+x)^2) is 9/10

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integral from 0 to 9 of 1/((1+x)^2)

\int_{0}^{9} \frac{1}{( 1 + x ) ^{2}} d x

\frac{9}{10}

Decimal Notation

0.9

Solution steps

\int_{0}^{9} \frac{1}{( 1 + x ) ^{2}} d x

Apply u-substitution

: \int_{1}^{10} \frac{1}{u ^{2}} d u

= \int_{1}^{10} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{1}^{10}

= [- \frac{1}{u}]_{1}^{10}

Compute the boundaries:

\frac{9}{10}

= \frac{9}{10}

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What is the integral from 0 to 9 of 1/((1+x)^2) ?
The integral from 0 to 9 of 1/((1+x)^2) is 9/10