What is the integral from-1 to 1 of (2x)/((x+7)^2) ?

The integral from-1 to 1 of (2x)/((x+7)^2) is 2(ln(4/3)-7/24)

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integral from-1 to 1 of (2x)/((x+7)^2)

\int_{- 1}^{1} \frac{2 x}{( x + 7 ) ^{2}} d x

2 (ln (\frac{4}{3}) - \frac{7}{24})

Decimal Notation

- 0.00796 \dots

Solution steps

\int_{- 1}^{1} \frac{2 x}{( x + 7 ) ^{2}} d x

Take the constant out:

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

= 2 \cdot \int_{- 1}^{1} \frac{x}{( x + 7 ) ^{2}} d x

Apply u-substitution

: \int_{6}^{8} \frac{u - 7}{u ^{2}} d u

= 2 \cdot \int_{6}^{8} \frac{u - 7}{u ^{2}} d u

Expand

\frac{u - 7}{u ^{2}} : \frac{1}{u} - \frac{7}{u ^{2}}

= 2 \cdot \int_{6}^{8} \frac{1}{u} - \frac{7}{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

= 2 (\int_{6}^{8} \frac{1}{u} d u - \int_{6}^{8} \frac{7}{u ^{2}} d u)

\int_{6}^{8} \frac{1}{u} d u = ln (\frac{4}{3})

\int_{6}^{8} \frac{7}{u ^{2}} d u = \frac{7}{24}

= 2 (ln (\frac{4}{3}) - \frac{7}{24})

\frac{\partial}{\partial z} (2 x yz)

6 x^{2} y^{^{'}} = y^{^{'}} + 3 x e^{- y}

\frac{d}{d x} (- \frac{1}{2} + \frac{7}{5} x^{3})

\frac{d}{d x} ((x^{3} - 2 x^{2} + 3) (7 x^{2} - 4 x))

\frac{d}{d y} (k y (1 - \frac{y}{N}) (\frac{y}{M} - 1))

What is the integral from-1 to 1 of (2x)/((x+7)^2) ?
The integral from-1 to 1 of (2x)/((x+7)^2) is 2(ln(4/3)-7/24)