What is the integral from 0 to 1 of x/((x+3)^2) ?

The integral from 0 to 1 of x/((x+3)^2) is 2ln(2)-ln(3)-1/4

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integral from 0 to 1 of x/((x+3)^2)

\int_{0}^{1} \frac{x}{( x + 3 ) ^{2}} d x

2 ln (2) - ln (3) - \frac{1}{4}

Decimal

0.03768 \dots

Solution steps

\int_{0}^{1} \frac{x}{( x + 3 ) ^{2}} d x

Apply u-substitution

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

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

Expand

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

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

\int_{3}^{4} \frac{1}{u} d u = 2 ln (2) - ln (3)

\int_{3}^{4} \frac{3}{u ^{2}} d u = \frac{1}{4}

= 2 ln (2) - ln (3) - \frac{1}{4}

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What is the integral from 0 to 1 of x/((x+3)^2) ?
The integral from 0 to 1 of x/((x+3)^2) is 2ln(2)-ln(3)-1/4