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

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

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integral from 0 to 1 of x/(x^2+4x+4)

\int_{0}^{1} \frac{x}{x ^{2} + 4 x + 4} d x

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

Decimal Notation

0.07213 \dots

Solution steps

\int_{0}^{1} \frac{x}{x ^{2} + 4 x + 4} d x

Complete the square

x^{2} + 4 x + 4 : (x + 2)^{2}

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

\frac{\partial}{\partial x} (e^{- 6 x^{2} - 3 y^{2}})

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