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

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

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integral of 1/(at+b)

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

\frac{1}{a} ln ∣ b + a t ∣ + C

Solution steps

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

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 = b + a t

= \frac{1}{a} ln ∣ b + a t ∣

Add a constant to the solution

= \frac{1}{a} ln ∣ b + a t ∣ + C

\frac{x}{x ^{2} + 121}

\frac{d}{d x} (2^{(5 x - 3 x^{2})})

(1, - 1), (- 1, - 2)

\int_{0}^{1} x e^{- x^{2}} d x

\int - \frac{1}{1 + x ^{2}} d x