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

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

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

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

ln ∣ x + 4 ∣ + \frac{4}{x + 4} + C

Solution steps

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

Apply u-substitution

: \int \frac{u - 4}{u ^{2}} d u

= \int \frac{u - 4}{u ^{2}} d u

Expand

\frac{u - 4}{u ^{2}} : \frac{1}{u} - \frac{4}{u ^{2}}

= \int \frac{1}{u} - \frac{4}{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{4}{u ^{2}} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{4}{u ^{2}} d u = - \frac{4}{u}

= ln ∣ u ∣ - (- \frac{4}{u})

Substitute back

u = x + 4

= ln ∣ x + 4 ∣ - (- \frac{4}{x + 4})

Simplify

= ln ∣ x + 4 ∣ + \frac{4}{x + 4}

Add a constant to the solution

= ln ∣ x + 4 ∣ + \frac{4}{x + 4} + C

\int \frac{x}{x + 3} d x

\frac{\partial}{\partial y} (x + e^{y})

t \to 3 lim (2 t^{- 1})

\frac{d}{d x} (\frac{5 + 6 x - 8 x}{x})

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

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