Question

Let g(x) be an antiderivative for f(x). Then In $(1+(g\left(x\right){)}^2)$ is an antiderivate 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 $\frac{2f\left(x\right)g\left(x\right)}{1+\left(g\right(x){)}^2}$

Given $\int f\left(x\right)dx=g\left(x\right)g^{\prime }\left(x\right)=f\left(x\right)$
Now $\frac{d}{dx}\left(In\right(1+g^2\left(x\right))=\frac{2g\left(x\right)g^{\prime }\left(x\right)}{1+g^2\left(x\right)}\frac{2f\left(x\right)g\left(x\right)}{1+g^2\left(x\right)}$