What is the integral of 4/(5-3x) ?

The integral of 4/(5-3x) is -4/3 ln|5-3x|+C

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integral of 4/(5-3x)

\int \frac{4}{5 - 3 x} d x

- \frac{4}{3} ln ∣ 5 - 3 x ∣ + C

Solution steps

\int \frac{4}{5 - 3 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 - 3 x} d x

Apply u-substitution

: \int - \frac{1}{3 u} d u

= 4 \cdot \int - \frac{1}{3 u} d u

Take the constant out:

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

= 4 (- \frac{1}{3} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 (- \frac{1}{3} ln ∣ u ∣)

Substitute back

u = 5 - 3 x

= 4 (- \frac{1}{3} ln ∣ 5 - 3 x ∣)

Simplify

4 (- \frac{1}{3} ln ∣ 5 - 3 x ∣) : - \frac{4}{3} ln ∣ 5 - 3 x ∣

= - \frac{4}{3} ln ∣ 5 - 3 x ∣

Add a constant to the solution

= - \frac{4}{3} ln ∣ 5 - 3 x ∣ + C

\frac{\partial}{\partial θ} (r cos (θ))

x^{\frac{1}{x}}

\frac{d y}{d t} = 2 (y^{2} - y) t

\int 5 x + 4 d x

y = x^{4} - 3 x^{2} + 10, at x = 1