What is the integral of (x+4)/(2x+6) ?

The integral of (x+4)/(2x+6) is 1/2 (x+3+ln(2|x+3|))+C

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integral of (x+4)/(2x+6)

\int \frac{x + 4}{2 x + 6} d x

\frac{1}{2} (x + 3 + ln (2 ∣ x + 3 ∣)) + C

Solution steps

\int \frac{x + 4}{2 x + 6} 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 x + 6

= \frac{1}{4} (2 x + 6 + 2 ln ∣ 2 x + 6 ∣)

Simplify

\frac{1}{4} (2 x + 6 + 2 ln ∣ 2 x + 6 ∣) : \frac{1}{2} (x + 3 + ln (2 ∣ x + 3 ∣))

= \frac{1}{2} (x + 3 + ln (2 ∣ x + 3 ∣))

Add a constant to the solution

= \frac{1}{2} (x + 3 + ln (2 ∣ x + 3 ∣)) + C

\int r e^{r} d r

2 x e^{x - 1} + e^{x - 1} (x^{2} + 1)

\int (e^{3 x} - 4)^{5} \cdot 3 e^{3 x} d x

\int arccot (9 y) d y

\frac{d ^{2}}{d x ^{2}} ((x^{3} + 8) e^{x})

What is the integral of (x+4)/(2x+6) ?
The integral of (x+4)/(2x+6) is 1/2 (x+3+ln(2|x+3|))+C