What is the integral from-1 to 1 of (5x)/((x+4)^2) ?

The integral from-1 to 1 of (5x)/((x+4)^2) is 5(ln(5)-ln(3)-8/15)

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integral from-1 to 1 of (5x)/((x+4)^2)

\int_{- 1}^{1} \frac{5 x}{( x + 4 ) ^{2}} d x

5 (ln (5) - ln (3) - \frac{8}{15})

Decimal Notation

- 0.11253 \dots

Solution steps

\int_{- 1}^{1} \frac{5 x}{( x + 4 ) ^{2}} d x

Take the constant out:

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

= 5 \cdot \int_{- 1}^{1} \frac{x}{( x + 4 ) ^{2}} d x

Apply u-substitution

: \int_{3}^{5} \frac{u - 4}{u ^{2}} d u

= 5 \cdot \int_{3}^{5} \frac{u - 4}{u ^{2}} d u

Expand

\frac{u - 4}{u ^{2}} : \frac{1}{u} - \frac{4}{u ^{2}}

= 5 \cdot \int_{3}^{5} \frac{1}{u} - \frac{4}{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

= 5 (\int_{3}^{5} \frac{1}{u} d u - \int_{3}^{5} \frac{4}{u ^{2}} d u)

\int_{3}^{5} \frac{1}{u} d u = ln (5) - ln (3)

\int_{3}^{5} \frac{4}{u ^{2}} d u = \frac{8}{15}

= 5 (ln (5) - ln (3) - \frac{8}{15})

y = ∣ x - 6 ∣, y = \frac{x}{2}

\int \frac{x + 9}{36 - ( x - 6 ) ^{2}} d x

n = 1 \sum \infty \frac{2 n}{5 n - 3}

e^{s i n (2 x)}

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

What is the integral from-1 to 1 of (5x)/((x+4)^2) ?
The integral from-1 to 1 of (5x)/((x+4)^2) is 5(ln(5)-ln(3)-8/15)