What is the integral of x/((1+2x)^2) ?

The integral of x/((1+2x)^2) is 1/4 (ln|1+2x|+1/(1+2x))+C

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integral of x/((1+2x)^2)

\int \frac{x}{( 1 + 2 x ) ^{2}} d x

\frac{1}{4} (ln ∣ 1 + 2 x ∣ + \frac{1}{1 + 2 x}) + C

Solution steps

\int \frac{x}{( 1 + 2 x ) ^{2}} d x

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Expand

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

= \frac{1}{4} \cdot \int \frac{1}{u} - \frac{1}{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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

= \frac{1}{4} (ln ∣ u ∣ - (- \frac{1}{u}))

Substitute back

u = 1 + 2 x

= \frac{1}{4} (ln ∣ 1 + 2 x ∣ - (- \frac{1}{1 + 2 x}))

Simplify

= \frac{1}{4} (ln ∣ 1 + 2 x ∣ + \frac{1}{1 + 2 x})

Add a constant to the solution

= \frac{1}{4} (ln ∣ 1 + 2 x ∣ + \frac{1}{1 + 2 x}) + C

\frac{\partial}{\partial t} (2 s^{2} t^{2})

\int x^{2} 8 + x d x

\frac{d}{d x} (arccot (\frac{1}{2}) (x - \frac{1}{x}))

\int \frac{7}{9 + 4 x ^{2}} d x

\frac{d}{d x} (ln (\frac{x}{7}))

What is the integral of x/((1+2x)^2) ?
The integral of x/((1+2x)^2) is 1/4 (ln|1+2x|+1/(1+2x))+C