What is the integral of 1/(4-3x) ?

The integral of 1/(4-3x) is -1/3 ln|4-3x|+C

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integral of 1/(4-3x)

\int \frac{1}{4 - 3 x} d x

- \frac{1}{3} ln ∣ 4 - 3 x ∣ + C

Solution steps

\int \frac{1}{4 - 3 x} d x

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{3} ln ∣ u ∣

Substitute back

u = 4 - 3 x

= - \frac{1}{3} ln ∣ 4 - 3 x ∣

Add a constant to the solution

= - \frac{1}{3} ln ∣ 4 - 3 x ∣ + C

(2.5) (- 1.7)

x \to 0 lim (\frac{x ^{4}}{x ^{2}})

n = 0 \sum \infty \frac{x ^{n}}{2 ^{n}}

n = 1 \sum \infty \frac{1}{2 ^{n} - n}

y^{^{''}} + \frac{y ^{^{'}}}{x} = 0