What is the integral of 5/(2-4x) ?

The integral of 5/(2-4x) is -5/4 ln|2-4x|+C

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integral of 5/(2-4x)

\int \frac{5}{2 - 4 x} d x

- \frac{5}{4} ln ∣ 2 - 4 x ∣ + C

Solution steps

\int \frac{5}{2 - 4 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}{2 - 4 x} d x

Apply u-substitution

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

= 5 \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

= 5 (- \frac{1}{4} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 (- \frac{1}{4} ln ∣ u ∣)

Substitute back

u = 2 - 4 x

= 5 (- \frac{1}{4} ln ∣ 2 - 4 x ∣)

Simplify

5 (- \frac{1}{4} ln ∣ 2 - 4 x ∣) : - \frac{5}{4} ln ∣ 2 - 4 x ∣

= - \frac{5}{4} ln ∣ 2 - 4 x ∣

Add a constant to the solution

= - \frac{5}{4} ln ∣ 2 - 4 x ∣ + C

\int 2 x + 3 x^{2} d x

y = \frac{x - 4}{x + 2}

\int \frac{5}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\int \frac{e ^{8 x}}{e ^{8 x} + 2} d x

\frac{2}{s ^{2} + 4}