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

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

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integral of 1/(8x+1)

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

\frac{1}{8} ln ∣ 8 x + 1 ∣ + C

Solution steps

\int \frac{1}{8 x + 1} 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 = 8 x + 1

= \frac{1}{8} ln ∣ 8 x + 1 ∣

Add a constant to the solution

= \frac{1}{8} ln ∣ 8 x + 1 ∣ + C

x \to 1 + lim (ln (x - 1))

\int e^{4 x + 7} d x

e^{x} \cdot sin (2 x)

f (t) = t^{2} + 6 t - 3

f (x) = 2 x^{4}, (- 2, 32)