Question
∫x{f(x2)g′′(x2)−f′′(x2)g(x2)}dx
Answer: Option C
:
C
Put x2=t⇒I=12∫{f(t)g′′(t)−g(t)f′′(t)}dt=12{f(t)g′(t)−∫f′(t)g′(t)−g(t)f′(t)+∫g′(t)f′(t)dt}+c∴I=12{f(t)g′(t)−g(t)f′(t)}+c=12{f(x2)g′(x2)−g(x2)f′(x2)}+c
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:
C
Put x2=t⇒I=12∫{f(t)g′′(t)−g(t)f′′(t)}dt=12{f(t)g′(t)−∫f′(t)g′(t)−g(t)f′(t)+∫g′(t)f′(t)dt}+c∴I=12{f(t)g′(t)−g(t)f′(t)}+c=12{f(x2)g′(x2)−g(x2)f′(x2)}+c
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