What is the integral of 1/(4+8x) ?

The integral of 1/(4+8x) is 1/8 ln|4+8x|+C

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integral of 1/(4+8x)

\int \frac{1}{4 + 8 x} d x

\frac{1}{8} ln ∣ 4 + 8 x ∣ + C

Solution steps

\int \frac{1}{4 + 8 x} d x

Apply u-substitution

: \int \frac{1}{8 u} d u

= \int \frac{1}{8 u} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{8} ln ∣ u ∣

Substitute back

u = 4 + 8 x

= \frac{1}{8} ln ∣ 4 + 8 x ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ 4 + 8 x ∣ + C

\int 7 x^{\frac{2}{5}} + 3 x^{- \frac{4}{5}} d x

y^{^{'''}} - 5 y^{^{''}} + 3 y^{^{'}} + 9 y = 0

x \to - 10 lim (- 15)

\frac{d}{d x} (x, π x)

\frac{10}{1 + e ^{- x}}