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

The integral from 0 to 1/2 of (2x+1)^{-2} is 1/4

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

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

\frac{1}{4}

Decimal

0.25

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2} \cdot \frac{1}{2}

Simplify

= \frac{1}{4}

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What is the integral from 0 to 1/2 of (2x+1)^{-2} ?
The integral from 0 to 1/2 of (2x+1)^{-2} is 1/4