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

The integral from 0 to 1 of 2/((x+2)^2) is 1/3

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

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

\frac{1}{3}

Decimal Notation

0.33333 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{2}^{3} \frac{1}{u ^{2}} d u

= 2 \cdot \int_{2}^{3} \frac{1}{u ^{2}} d u

Apply the Power Rule

: [- \frac{1}{u}]_{2}^{3}

= 2 [- \frac{1}{u}]_{2}^{3}

Compute the boundaries:

\frac{1}{6}

= 2 \cdot \frac{1}{6}

Simplify

= \frac{1}{3}

\frac{d}{d x} (\frac{1 ( e ^{x} - e ^{- x} )}{2})

\int \frac{- 2}{5 x ^{2}} d x

\frac{d y}{d x} - x y^{2} = 2 y x

\frac{d}{d x} (ln (\frac{3 - x}{3 + x}))

\frac{d}{d t} (0.5 t^{2} e^{- 0.6 t})

What is the integral from 0 to 1 of 2/((x+2)^2) ?
The integral from 0 to 1 of 2/((x+2)^2) is 1/3