What is the integral from-1 to 1 of (4x)/((x+3)^2) ?

The integral from-1 to 1 of (4x)/((x+3)^2) is 4ln(2)-3

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integral from-1 to 1 of (4x)/((x+3)^2)

\int_{- 1}^{1} \frac{4 x}{( x + 3 ) ^{2}} d x

4 ln (2) - 3

Decimal Notation

- 0.22741 \dots

Solution steps

\int_{- 1}^{1} \frac{4 x}{( x + 3 ) ^{2}} d x

Take the constant out:

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

= 4 \cdot \int_{- 1}^{1} \frac{x}{( x + 3 ) ^{2}} d x

Apply u-substitution

: \int_{2}^{4} \frac{u - 3}{u ^{2}} d u

= 4 \cdot \int_{2}^{4} \frac{u - 3}{u ^{2}} d u

Expand

\frac{u - 3}{u ^{2}} : \frac{1}{u} - \frac{3}{u ^{2}}

= 4 \cdot \int_{2}^{4} \frac{1}{u} - \frac{3}{u ^{2}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int_{2}^{4} \frac{1}{u} d u - \int_{2}^{4} \frac{3}{u ^{2}} d u)

\int_{2}^{4} \frac{1}{u} d u = ln (2)

\int_{2}^{4} \frac{3}{u ^{2}} d u = \frac{3}{4}

= 4 (ln (2) - \frac{3}{4})

Simplify

4 (ln (2) - \frac{3}{4}) : 4 ln (2) - 3

= 4 ln (2) - 3

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x \to - 3 lim (\frac{- 6 x - 18}{x ^{3} + 3 x ^{2}})

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What is the integral from-1 to 1 of (4x)/((x+3)^2) ?
The integral from-1 to 1 of (4x)/((x+3)^2) is 4ln(2)-3