Question

Find a function $f$that has the given derivative $f^{\text{'}}$and value f(c).

$f\text{'}\left(x\right)=\frac{x}{1+4x^2},f\left(0\right)=1$

Short answer

The anti derivative is $f\left(x\right)=\frac{1}{8}\ln \left(1+4x^2\right)+1$

Step-by-step solution

Step 1:Given information

The given expression is$f\text{'}\left(x\right)=\frac{x}{1+4x^2}$

Step 2:Simplification

Given

$f\text{'}\left(x\right)=\frac{x}{1+4x^2}$

We will use an intelligent guess-and-check method to find f

Since differentiating a power function one and derivative of $\ln xis\frac{1}{x}$

One can start with $f\left(x\right)=\ln \left(1+4x^2\right)$

Now,

$f\text{'}\left(x\right)=\frac{d}{dx}\left\{\ln \left(1+4x^2\right)\right\}$

$=\frac{1}{1+4x^2}\frac{d}{dx}\left(1+4x^2\right)$

$=\frac{8x}{1+4x^2}$

Hence, we can guess the function as

$f\left(x\right)=\frac{1}{8}\ln \left(1+4x^2\right)+c$

Now ,

$$\begin{gathered} f\left(0\right)=1 \\ \Rightarrow 1=\ln (1+0)+c \end{gathered}$$

$\Rightarrow c=1$

Therefore,

$f\left(x\right)=\frac{1}{8}\ln \left(1+4x^2\right)+1$