What is the integral of 1/((ax+b)) ?

The integral of 1/((ax+b)) is 1/a ln|ax+b|+C

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integral of 1/((ax+b))

\int \frac{1}{( a x + b )} d x

\frac{1}{a} ln ∣ a x + b ∣ + C

Solution steps

\int \frac{1}{a x + b} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{a} ln ∣ u ∣

Substitute back

u = a x + b

= \frac{1}{a} ln ∣ a x + b ∣

Add a constant to the solution

= \frac{1}{a} ln ∣ a x + b ∣ + C

e^{x^{2}} + 5 x^{2}

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

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

\frac{d}{d x} (ln cos (\frac{π}{x}))

\int_{0}^{4} 5 d x

What is the integral of 1/((ax+b)) ?
The integral of 1/((ax+b)) is 1/a ln|ax+b|+C