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

The integral from-1 to 1 of (6x)/((x+6)^2) is 6(ln(7)-ln(5)-12/35)

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

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

6 (ln (7) - ln (5) - \frac{12}{35})

Decimal Notation

- 0.03830 \dots

Solution steps

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

Apply u-substitution

: \int_{5}^{7} \frac{u - 6}{u ^{2}} d u

= 6 \cdot \int_{5}^{7} \frac{u - 6}{u ^{2}} d u

Expand

\frac{u - 6}{u ^{2}} : \frac{1}{u} - \frac{6}{u ^{2}}

= 6 \cdot \int_{5}^{7} \frac{1}{u} - \frac{6}{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_{5}^{7} \frac{1}{u} d u - \int_{5}^{7} \frac{6}{u ^{2}} d u)

\int_{5}^{7} \frac{1}{u} d u = ln (7) - ln (5)

\int_{5}^{7} \frac{6}{u ^{2}} d u = \frac{12}{35}

= 6 (ln (7) - ln (5) - \frac{12}{35})

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What is the integral from-1 to 1 of (6x)/((x+6)^2) ?
The integral from-1 to 1 of (6x)/((x+6)^2) is 6(ln(7)-ln(5)-12/35)