Q. $\int_{}^{}{x\{ f\left( x^{2} \right)g^{''}\left( x^{2} \right) - f^{''}\left( x^{2} \right)g\left( x^{2} \right)\} dx}$ is equal to

A) $f\left( x^{2} \right)g^{'}\left( x^{2} \right) - g\left( x^{2} \right)f^{'}\left( x^{2} \right) + c$
B) $\frac{1}{2}\left\{ f\left( x^{2} \right)g\left( x^{2} \right)f^{'}\left( x^{2} \right) \right\} + c$
C) $\frac{1}{2}\left\{ f\left( x^{2} \right)g^{'}\left( x^{2} \right) - g\left( x^{2} \right)f^{'}\left( x^{2} \right) \right\} + c$
D) None of the above
Explanation

(c)
Let $I = \int_{}^{}{x\{ f\left( x^{2} \right)g^{''}\left( x^{2} \right) - f^{''}\left( x^{2} \right)g\left( x^{2} \right)\} dx}$
Put $x^{2} = t\ \Rightarrow 2x\ dx = dt$
$\therefore I = \frac{1}{2}\lbrack\ \left\{ f(t)g^{''}(t) \right\} dt - \int_{}^{}{f^{''}\left( t^{2} \right)g\left( t^{2} \right)dt\rbrack\ }$
$= \frac{1}{2}\lbrack\ f(t)g^{'}(t) - \int_{}^{}{f^{'}(t)g(t)dt - g(t)f^{'}(t) + \int_{}^{}{g^{'}(t)f^{'}(t)dt\ \rbrack + c}}$
$= \frac{1}{2}\left\lbrack \ f(t)g^{'}(t) - g(t)f^{'}(t) \right\rbrack + c$
$= \frac{1}{2}\left\lbrack \ f\left( x^{2} \right)g^{'}\left( x^{2} \right) - g\left( x^{2} \right)f^{'}\left( x^{2} \right) \right\rbrack + c$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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