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

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

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integral from-1 to 1 of (3x)/((x+5)^2)

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

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

Decimal Notation

- 0.03360 \dots

Solution steps

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

Take the constant out:

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

= 3 \cdot \int_{- 1}^{1} \frac{x}{( x + 5 ) ^{2}} d x

Apply u-substitution

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

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

Expand

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

= 3 \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

= 3 (\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}

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

f (x) = - x^{2} - 4 x - 6, at x = 4

\int cos (\frac{π t}{2}) d t

\int \frac{tan ^{3} ( ln ( x ) )}{8 x} d x

\frac{d ^{2}}{d x ^{2}} y = sin (x)

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