What is the integral of 1/(1+16x) ?

The integral of 1/(1+16x) is 1/16 ln|1+16x|+C

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integral of 1/(1+16x)

\int \frac{1}{1 + 16 x} d x

\frac{1}{16} ln ∣ 1 + 16 x ∣ + C

Solution steps

\int \frac{1}{1 + 16 x} d x

Apply u-substitution

: \int \frac{1}{16 u} d u

= \int \frac{1}{16 u} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{16} ln ∣ u ∣

Substitute back

u = 1 + 16 x

= \frac{1}{16} ln ∣ 1 + 16 x ∣

Add a constant to the solution

= \frac{1}{16} ln ∣ 1 + 16 x ∣ + C

\frac{d y}{d x} - \frac{y}{x} = 7 x e^{x}

\int_{1}^{36} \frac{1}{2 x} d x

x \to - 5 lim (3 x + 2)

\int x^{3} 81 + x^{2} d x

\frac{\partial}{\partial x} (e^{- \frac{x}{2 y}})

What is the integral of 1/(1+16x) ?
The integral of 1/(1+16x) is 1/16 ln|1+16x|+C