What is the integral of 4/(5+4x) ?

The integral of 4/(5+4x) is ln|5+4x|+C

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integral of 4/(5+4x)

\int \frac{4}{5 + 4 x} d x

ln ∣ 5 + 4 x ∣ + C

Solution steps

\int \frac{4}{5 + 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}{5 + 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 = 5 + 4 x

= 4 \cdot \frac{1}{4} ln ∣ 5 + 4 x ∣

Simplify

4 \cdot \frac{1}{4} ln ∣ 5 + 4 x ∣ : ln ∣ 5 + 4 x ∣

= ln ∣ 5 + 4 x ∣

Add a constant to the solution

= ln ∣ 5 + 4 x ∣ + C

\int 3 sin (y) + 1 d y

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

(e^{x} ln (y)) d x + (\frac{e ^{2 x}}{2 y}) d y = 0, y (0) = e

h (x) = \frac{6}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}}

y^{^{''}} - 4 y = 0, y (2) = 2, y^{^{'}} (- 1) = 1