What is the integral of 1/((4+x)) ?

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

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integral of 1/((4+x))

integral

\frac{1}{( 4 + x )}

ln ∣ 4 + x ∣ + C

Solution steps

\int \frac{1}{4 + x} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = 4 + x

= ln ∣ 4 + x ∣

Add a constant to the solution

= ln ∣ 4 + x ∣ + C

f (x) = \frac{1}{x ^{\frac{1}{2}}}

\frac{x + 1}{3 x + 2}

sin^{6} (x) cos^{4} (x)

\frac{( 1 - cos ( x ) ) ^{2}}{sin ^{2} ( x )}

3^{- x}