# Thread: Find antiderivative here

1. ## Find antiderivative here

Is antiderivative of $\displaystyle (6x^2+1)$ is it $\displaystyle 2x^3+x+C$?
Is antiderivative of $\displaystyle \frac{1}{4z}$ is it $\displaystyle \frac{1}{4}\ln4z+C$?

2. Originally Posted by Critter314
Is antiderivative of $\displaystyle (6x^2+1)$ is it $\displaystyle 2x^3+x+C$?
Yes (Clap)

Is antiderivative of $\displaystyle \frac{1}{4z}$ is it $\displaystyle \frac{1}{4}\ln4z+C$?
Yes (Clap)

3. Hello, Critter314!

$\displaystyle \text{The antiderivative of }\,\dfrac{1}{4z}$

$\displaystyle \text{Is it }\,\frac{1}{4}\ln4z+C\,?$

Well, yes and no . . .

Note that: .$\displaystyle \displaystyle \int\frac{dz}{4z} \;=\;\tfrac{1}{4}\int\frac{dz}{z} \;=\; \tfrac{1}{4}\ln z + C$

Your answer is correct, but it can be simplified.

. . $\displaystyle \frac{1}{4}\ln4z + C \;=\; \frac{1}{4}\bigg[\log 4 + \ln z\bigg] + C$

. . . . . . . . . . .$\displaystyle =\;\underbrace{\tfrac{1}{4}\ln4}_{\text{constant}} + \frac{1}{4}\ln z + C$

. . . . . . . . . . .$\displaystyle =\;\tfrac{1}{4}\ln z + C$