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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

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

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

Compute the boundaries:

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

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

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