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

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

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integral of 5/(4-3x)

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

- \frac{5}{3} ln ∣ 4 - 3 x ∣ + C

Solution steps

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

Take the constant out:

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

= 5 \cdot \int \frac{1}{4 - 3 x} d x

Apply u-substitution

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

= 5 \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

= 5 (- \frac{1}{3} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 (- \frac{1}{3} ln ∣ u ∣)

Substitute back

u = 4 - 3 x

= 5 (- \frac{1}{3} ln ∣ 4 - 3 x ∣)

Simplify

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

= - \frac{5}{3} ln ∣ 4 - 3 x ∣

Add a constant to the solution

= - \frac{5}{3} ln ∣ 4 - 3 x ∣ + C

\int (ln (x))^{2} \frac{1}{x ^{2}} d x

\frac{\partial}{\partial x} (sin (5 x y + 4 yz + 7 x z))

f (x) = \frac{- 8 x}{x ^{2} + 1}, (0, 0)

\int_{2}^{5} x d x

\int e^{- \frac{1}{2} z^{2}}