What is the integral of 4/(3+4x) ?

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

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integral of 4/(3+4x)

\int \frac{4}{3 + 4 x} d x

ln ∣ 3 + 4 x ∣ + C

Solution steps

\int \frac{4}{3 + 4 x} 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}{3 + 4 x} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = 3 + 4 x

= 4 \cdot \frac{1}{4} ln ∣ 3 + 4 x ∣

Simplify

4 \cdot \frac{1}{4} ln ∣ 3 + 4 x ∣ : ln ∣ 3 + 4 x ∣

= ln ∣ 3 + 4 x ∣

Add a constant to the solution

= ln ∣ 3 + 4 x ∣ + C

\int_{1}^{49} \frac{1}{2 x} d x

\frac{d}{d x} (\frac{x ^{2}}{3})

t \cdot e^{- 5 t}

f (x) = 5 x + 4

x \to - 7 lim (\frac{1}{x ^{2} ( x + 7 )})