∫ x { f ( x 2 ) g ′′ ( x 2 ) − f ′′ ( x 2 ) g ( x 2 ) } d x Options: A . &nbsp f ( x 2 ) g ′ ( x 2 ) − g ( x 2 ) f ′ ( x 2 ) + c B . &nbsp 1 2 { f ( x 2 ) g ( x 2 ) f ′ ( x 2 ) } + c C . &nbsp 1 2 { f ( x 2 ) g ′ ( x 2 ) − g ( x 2 ) f ′ ( x 2 ) } + c D . &nbsp None of these

Answer: Option C : C Put x 2 = t ⇒ I = 1 2 ∫ { f ( t ) g ′′ ( t ) − g ( t ) f ′′ ( t ) } d t = 1 2 { f ( t ) g ′ ( t ) − ∫ f ′ ( t ) g ′ ( t ) − g ( t ) f ′ ( t ) + ∫ g ′ ( t ) f ′ ( t ) d t } + c ∴ I = 1 2 { f ( t ) g ′ ( t ) − g ( t ) f ′ ( t ) } + c = 1 2 { f ( x 2 ) g ′ ( x 2 ) − g ( x 2 ) f ′ ( x 2 ) } + c

∫x{f(x2)g′′(x2)−f′′(x2)g(x2)}dx

Discussion Forum : Indefinite Integration

Question -

$\int x {f (x^{2}) g^{''} (x^{2}) - f^{''} (x^{2}) g (x^{2})} d x$

Options:

A .

f (x^{2}) g^{^{'}} (x^{2}) - g (x^{2}) f^{^{'}} (x^{2}) + c

B .

\frac{1}{2} {f (x^{2}) g (x^{2}) f^{^{'}} (x^{2})} + c

C .

\frac{1}{2} {f (x^{2}) g^{^{'}} (x^{2}) - g (x^{2}) f^{^{'}} (x^{2})} + c

D . None of these

Answer: Option C
:
C
Put

x^{2} = t \Rightarrow I = \frac{1}{2} \int {f (t) g^{''} (t) - g (t) f^{''} (t)} d t = \frac{1}{2} {f (t) g^{^{'}} (t) - \int f^{^{'}} (t) g^{^{'}} (t) - g (t) f^{^{'}} (t) + \int g^{^{'}} (t) f^{^{'}} (t) d t} + c ∴ I = \frac{1}{2} {f (t) g^{^{'}} (t) - g (t) f^{^{'}} (t)} + c = \frac{1}{2} {f (x^{2}) g^{^{'}} (x^{2}) - g (x^{2}) f^{^{'}} (x^{2})} + c

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Discussion Forum : Indefinite Integration

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