If $f(x)$ is anti-derivative of $g(x)$ and $\int f(x) g(x)\left(1+f^2(x)\right) d x=F(x)$, then $F(x)=$

If $f(x)$ is anti-derivative of $g(x)$ and $\int f(x) g(x)\left(1+f^2(x)\right) d x=F(x)$, then $F(x)=$
  1. $\frac{\left.\left(1+f^2 x\right)\right)^2}{4}+C$
  2. $\frac{\left.\left(1+f^2 x\right)\right)^2}{2}+C$
  3. $\frac{f^2(x) g(x)}{4}+C$
  4. $\frac{g^2(x) f(x)}{4}+C$

Solution

Given, $f(x)=\int g(x) d x$ $\Rightarrow f^{\prime}(x)=g(x)$ $F(x)=\int f(x) g(x)\left(1+f^2(x)\right) d x$ On putting $1+f^2(x)=t$ $\Rightarrow \quad 2 f(x): f^{\prime}(x) d x=d t$ $\Rightarrow \quad f(x) \cdot g(x) d x=\frac{d t}{2} \quad\left[\because f^{\prime}(x)=g(x)\right]$ $f(x)=\int t \cdot \frac{d t}{2}=\frac{1}{2} \cdot \frac{t^2}{2}+C=\frac{\left(1+f^2(x)\right)^2}{4}+C$

Asked in: AP EAMCET 2022 (07 Jul Shift 1)

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