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

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

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

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: [- \frac{1}{u}]_{1}^{2}

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

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

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\int_{0}^{2} x ln (8 x) d x

\int_{0}^{2} x^{2} - 3 x + 2 d x

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