If f(x) is an anti-derivative of g(x) and int | vedclass

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(A) Given that $f(x)$ is an anti-derivative of $g(x)$,we have $f'(x) = g(x)$.We need to evaluate the integral $F(x) = \int f(x) g(x) (1 + f^2(x)) dx$.

Vedclass · Accepted Answer

$\frac{(1 + f^2(x))^2}{4} + C$

Vedclass · Answer

(A) Given that $f(x)$ is an anti-derivative of $g(x)$,we have $f'(x) = g(x)$.We need to evaluate the integral $F(x) = \int f(x) g(x) (1 + f^2(x)) dx$.Let $u = 1 + f^2(x)$.Then,differentiating with respect to $x$,we get $du = 2 f(x) f'(x) dx$.Since $f'(x) = g(x)$,we have $du = 2 f(x) g(x) dx$,which implies $f(x) g(x) dx = \frac{du}{2}$.Substituting these into the integral,we get $F(x) = \int u \cdot \frac{du}{2} = \frac{1}{2} \int u du$.Integrating,we get $F(x) = \frac{1}{2} \cdot \frac{u^2}{2} + C = \frac{u^2}{4} + C$.Substituting back $u = 1 + f^2(x)$,we get $F(x) = \frac{(1 + f^2(x))^2}{4} + C$.

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

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