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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 $F(x)=$
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Verified Answer
The correct answer is:
$\frac{\left.\left(1+f^2 x\right)\right)^2}{4}+C$
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 $1+f^2(x)=t$
$\Rightarrow \quad 2 f(x): f^{\prime}(x) d x=d t$
$\Rightarrow \quad f(x) \cdot g(x) d x=\frac{d t}{2} \quad\left[\because f^{\prime}(x)=g(x)\right]$
$f(x)=\int t \cdot \frac{d t}{2}=\frac{1}{2} \cdot \frac{t^2}{2}+C=\frac{\left(1+f^2(x)\right)^2}{4}+C$
$\Rightarrow f^{\prime}(x)=g(x)$
$F(x)=\int f(x) g(x)\left(1+f^2(x)\right) d x$
On putting $1+f^2(x)=t$
$\Rightarrow \quad 2 f(x): f^{\prime}(x) d x=d t$
$\Rightarrow \quad f(x) \cdot g(x) d x=\frac{d t}{2} \quad\left[\because f^{\prime}(x)=g(x)\right]$
$f(x)=\int t \cdot \frac{d t}{2}=\frac{1}{2} \cdot \frac{t^2}{2}+C=\frac{\left(1+f^2(x)\right)^2}{4}+C$
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