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

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

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

integral

\frac{x}{( 2 + 2 x )}

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

\int 1 d u = u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= \frac{1}{4} (u - 2 ln ∣ u ∣)

Substitute back

u = 2 + 2 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{1}{e ^{x} + 1}

\frac{( 2 x + 5 )}{( x + 2 )}

sin (2 x + 1)

x + 16

\frac{1}{( 35 - x )}

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