What is the integral of 1/(3-8x) ?

The integral of 1/(3-8x) is -1/8 ln|3-8x|+C

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integral of 1/(3-8x)

\int \frac{1}{3 - 8 x} d x

- \frac{1}{8} ln ∣ 3 - 8 x ∣ + C

Solution steps

\int \frac{1}{3 - 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 = 3 - 8 x

= - \frac{1}{8} ln ∣ 3 - 8 x ∣

Add a constant to the solution

= - \frac{1}{8} ln ∣ 3 - 8 x ∣ + C

- \frac{x}{( 25 - x ^{2} ) ^{\frac{1}{2}}}

(125 + x)^{- \frac{1}{3}}

y^{^{''}} + 2 y^{^{'}} + y = 14 e^{- t}

\int cos^{3} (3 x) d x

y^{^{'}} = \frac{y}{( x ^{2} + 1 ) x}