Question

Let g(x) be an anti-derivative for f(x). Then $ln(1+(g\left(x\right){)}^2)$ is an antiderivative for

• A. $\frac{2f\left(x\right)g\left(x\right)}{1+\left(f\right(x){)}^2}$
• B. $\frac{2f\left(x\right)g\left(x\right)}{1+\left(g\right(x){)}^2}$
• C. $\frac{2f\left(x\right)}{1+\left(f\right(x){)}^2}$
• D. $\frac{2g\left(x\right)}{1+\left(g\right(x){)}^2}$

Answer 1

The correct option is B $\frac{2f\left(x\right)g\left(x\right)}{1+\left(g\right(x){)}^2}$

Given $\int f\left(x\right)dx=g\left(x\right)$
$\Rightarrow g^{\prime }\left(x\right)=f\left(x\right)$
Now, $\frac{d}{dx}ln[1+(g\left(x\right){)}^2]=\frac{2g\left(x\right).g^{\prime }\left(x\right)}{1+\left(g\right(x){)}^2}$
$=\frac{2f\left(x\right).g\left(x\right)}{1+\left(g\right(x){)}^2}$