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

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

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

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

5 (ln (2) - \frac{3}{4})

Decimal

- 0.28426 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{u - 3}{u ^{2}} : \frac{1}{u} - \frac{3}{u ^{2}}

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

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

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

= 5 (ln (2) - \frac{3}{4})

\int \frac{1}{1 + 4 y ^{2}} d y

x \to - 4 lim (7^{x})

\int_{2}^{6} ln (5 x) d x

\int 2 x ln (x^{2} + 5) d x

(4 e^{4 x})^{^{'}}

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