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

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

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

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

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

Decimal Notation

- 0.05685 \dots

Solution steps

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

Apply u-substitution

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

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

Expand

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

= \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

= \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}

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

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