What is the integral of 1/(7-4x) ?

The integral of 1/(7-4x) is -1/4 ln|7-4x|+C

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integral of 1/(7-4x)

\int \frac{1}{7 - 4 x} d x

- \frac{1}{4} ln ∣ 7 - 4 x ∣ + C

Solution steps

\int \frac{1}{7 - 4 x} d x

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{4} ln ∣ u ∣

Substitute back

u = 7 - 4 x

= - \frac{1}{4} ln ∣ 7 - 4 x ∣

Add a constant to the solution

= - \frac{1}{4} ln ∣ 7 - 4 x ∣ + C

y = e^{- 11 x}

n = 1 \sum \infty n^{3}

\frac{d x}{d t} = 3 t^{2}

\int \frac{1}{1 - 3 t} d t

y^{^{'}} + 7 y = t + e^{- 5 t}