What is the integral of 1/(6+2x) ?

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

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integral of 1/(6+2x)

\int \frac{1}{6 + 2 x} d x

\frac{1}{2} ln ∣ 6 + 2 x ∣ + C

Solution steps

\int \frac{1}{6 + 2 x} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 6 + 2 x

= \frac{1}{2} ln ∣ 6 + 2 x ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 6 + 2 x ∣ + C

\frac{\partial}{\partial x} (cos (x y) sin (2 y))

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

x - 2 ln (x + 1)

\int 8 x (ln (x^{7}))^{2} d x

y^{^{'}} + 2 y = e^{- t}