What is the integral from-1 to 1 of (6x)/((x+4)^2) ?

The integral from-1 to 1 of (6x)/((x+4)^2) is 6(ln(5)-ln(3)-8/15)

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integral from-1 to 1 of (6x)/((x+4)^2)

\int_{- 1}^{1} \frac{6 x}{( x + 4 ) ^{2}} d x

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

Solution steps

\int_{- 1}^{1} \frac{6 x}{( x + 4 ) ^{2}} d x

Take the constant out:

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

= 6 \cdot \int_{- 1}^{1} \frac{x}{( x + 4 ) ^{2}} d x

Apply u-substitution

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

= 6 \cdot \int_{3}^{5} \frac{u - 4}{u ^{2}} d u

Expand

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

= 6 \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

= 6 (\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}

= 6 (ln (5) - ln (3) - \frac{8}{15})

n = 0 \sum \infty n (z - 1)^{n - 1}

\frac{\partial}{\partial x} (\frac{82}{5 + x ^{2} + y ^{2}})

\int \frac{2}{s} d s

x \to - 2 lim (\frac{x}{x - 2})

\int_{1}^{3} e^{x + 2} d x

What is the integral from-1 to 1 of (6x)/((x+4)^2) ?
The integral from-1 to 1 of (6x)/((x+4)^2) is 6(ln(5)-ln(3)-8/15)