The antiderivative of 1/(ax+b) is 1/a ln|ax+b| + C where ln|x| = log_{e}|x| and C is a constant. In this post, we will learn how to find the antiderivative of 1/(ax+b), that is, how to integrate 1 divided by ax+b.

The antiderivative formula of 1/(ax+b), that is, the integral of 1/(ax+b) is given by

∫1/(ax+b) dx = 1/a ln|ax+b| + C,

where C is an integration constant.

## Antiderivative of 1/(ax+b)

The antiderivative of 1/(ax+b) is a function whose derivative is 1/(ax+b). Thus, the antiderivative of 1/(ax+b) equals ∫ 1/(ax+b) dx.

Now,

∫ $\dfrac{1}{ax+b}$ dx.

Let ax+b = t

Differentiating with respect to x,

a = dt/dx

⇒ dx = dt/a

∴ ∫ $\dfrac{1}{ax+b}$ dx = ∫ $\dfrac{dt}{a t}$

= $\dfrac{1}{a}$ ∫ $\dfrac{dt}{ t}$

= $\dfrac{1}{a}$ ln|t| + C

= $\dfrac{\ln |ax+b|}{a}$ + C as t=ax+b.

Hence, ∫ $\dfrac{1}{ax+b}$ dx = $\dfrac{\ln |ax+b|}{a}$ + C.

So the antiderivative of 1/(ax+b) is equal to 1/a ln|ax+b| +C where C is an integral constant.

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## FAQs

### Q1: What is the antiderivative of 1/(ax+b)?

Answer: The antiderivative of 1/(ax+b) is 1/a ln|ax+b| + C where C is an integration constant.

### Q2: What is the integral of 1/(ax+b)?

Answer: The integral of 1/(ax+b) is equal to 1/a ln|ax+b| + C.