What is the integral from 0 to 4 of (4-x)/((x+2)^2) ?

The integral from 0 to 4 of (4-x)/((x+2)^2) is -ln(3)+2

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integral from 0 to 4 of (4-x)/((x+2)^2)

\int_{0}^{4} \frac{4 - x}{( x + 2 ) ^{2}} d x

- ln (3) + 2

Decimal Notation

0.90138 \dots

Solution steps

\int_{0}^{4} \frac{4 - x}{( x + 2 ) ^{2}} d x

Expand

\frac{4 - x}{( x + 2 ) ^{2}} : \frac{4}{( x + 2 ) ^{2}} - \frac{x}{( x + 2 ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{4} \frac{4}{( x + 2 ) ^{2}} d x - \int_{0}^{4} \frac{x}{( x + 2 ) ^{2}} d x

\int_{0}^{4} \frac{4}{( x + 2 ) ^{2}} d x = \frac{4}{3}

\int_{0}^{4} \frac{x}{( x + 2 ) ^{2}} d x = ln (3) - \frac{2}{3}

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

Simplify

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

= - ln (3) + 2

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What is the integral from 0 to 4 of (4-x)/((x+2)^2) ?
The integral from 0 to 4 of (4-x)/((x+2)^2) is -ln(3)+2