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

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

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integral of 4/((x+1))

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

4 ln ∣ x + 1 ∣ + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 ln ∣ u ∣

Substitute back

u = x + 1

= 4 ln ∣ x + 1 ∣

Add a constant to the solution

= 4 ln ∣ x + 1 ∣ + C

\frac{d ^{2} y}{d t ^{2}} + 5 \frac{d y}{d t} + 6 y = 3 cos (2 t)

\frac{3}{( s + 2 )}

y^{^{''}} + 2 y^{^{'}} + 2 y = e^{- t} cos (t)

\frac{d}{d x} (ln (e^{s i n (x)} \cdot x^{c o s (x)}))

x \to - 1 lim (x^{3} + 3 x^{2} - 2 x - 3)