What is the integral from-1 to 1 of (2x)/((x+5)^2) ?

The integral from-1 to 1 of (2x)/((x+5)^2) is 2(ln(3/2)-5/12)

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integral from-1 to 1 of (2x)/((x+5)^2)

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

2 (ln (\frac{3}{2}) - \frac{5}{12})

Decimal Notation

- 0.02240 \dots

Solution steps

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

Apply u-substitution

: \int_{4}^{6} \frac{u - 5}{u ^{2}} d u

= 2 \cdot \int_{4}^{6} \frac{u - 5}{u ^{2}} d u

Expand

\frac{u - 5}{u ^{2}} : \frac{1}{u} - \frac{5}{u ^{2}}

= 2 \cdot \int_{4}^{6} \frac{1}{u} - \frac{5}{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_{4}^{6} \frac{1}{u} d u - \int_{4}^{6} \frac{5}{u ^{2}} d u)

\int_{4}^{6} \frac{1}{u} d u = ln (\frac{3}{2})

\int_{4}^{6} \frac{5}{u ^{2}} d u = \frac{5}{12}

= 2 (ln (\frac{3}{2}) - \frac{5}{12})

x \to 0 - lim (0)

3 t^{5}

\int (\frac{( x + 1 ) ( x - 2 )}{x}) d x

\int_{1}^{2} 4^{- θ} d θ

f (x) = 2 (2)^{4} - 16 (2)^{3} + 8

What is the integral from-1 to 1 of (2x)/((x+5)^2) ?
The integral from-1 to 1 of (2x)/((x+5)^2) is 2(ln(3/2)-5/12)