Question

Let $g\left(x\right)$ be an antiderivative for $f\left(x\right)$. 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. none

Answer 1

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

$g\left(x\right)$ is an antiderivative of $f\left(x\right)$

$\int f\left(x\right)=g\left(x\right)$ or $g^{\prime }\left(x\right)=f\left(x\right)$

Antiderivative of $\ln (1+(g\left(x\right){)}^2)$

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

$=\frac{2g\left(x\right)g^{\prime }\left(x\right)}{(1+(g\left(x\right){)}^2)}$

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