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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

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

Compute the boundaries:

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

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

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