What is the integral of x/(x^2+4x+4) ?

The integral of x/(x^2+4x+4) is ln|x+2|+2/(x+2)+C

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integral of x/(x^2+4x+4)

\int \frac{x}{x ^{2} + 4 x + 4} d x

ln ∣ x + 2 ∣ + \frac{2}{x + 2} + C

Solution steps

\int \frac{x}{x ^{2} + 4 x + 4} d x

Complete the square

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

= \int \frac{x}{( x + 2 ) ^{2}} d x

Apply u-substitution

: \int \frac{u - 2}{u ^{2}} d u

= \int \frac{u - 2}{u ^{2}} d u

Expand

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

= \int \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 \frac{1}{u} d u - \int \frac{2}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{2}{u ^{2}} d u = - \frac{2}{u}

= ln ∣ u ∣ - (- \frac{2}{u})

Substitute back

u = x + 2

= ln ∣ x + 2 ∣ - (- \frac{2}{x + 2})

Simplify

= ln ∣ x + 2 ∣ + \frac{2}{x + 2}

Add a constant to the solution

= ln ∣ x + 2 ∣ + \frac{2}{x + 2} + C

\frac{d}{d x} (2 x e^{2 x} + 2 x^{2} e^{2 x})

\int a sin (x) d x

\frac{d}{d x} (cos (x^{2} + y^{2}))

y^{2} = 4 x, y = 2 x + 4

y = 3 - x, y = 9 - x^{2}

What is the integral of x/(x^2+4x+4) ?
The integral of x/(x^2+4x+4) is ln|x+2|+2/(x+2)+C