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B
12{f(x2)g(x2)f′(x2)}+c
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C
12{f(x2)g′(x2)−g(x2)f′(x2)}+c
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D
None of these
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Solution
The correct option is C12{f(x2)g′(x2)−g(x2)f′(x2)}+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