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

Answer 1

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

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