If ∫x4+1x6+1dx=tan−1f(x)−23tan−1g(x)+c, where c is an arbitrary constant, then
A
both f(x) and g(x) are odd functions
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B
both f(x) and g(x) are even functions
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C
f(x)=g(x) has no real roots
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D
∫f(x)g(x)dx=−1x+13x3+d, where d is an arbitrary constant.
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Solution
The correct option is D∫f(x)g(x)dx=−1x+13x3+d, where d is an arbitrary constant. Let I=∫x4+1x6+1dx I=∫(x2+1)2−2x2(x2+1)(x4−x2+1)dx =∫(x2+1)(x4−x2+1)dx−2∫x2(x6+1)dx =∫(1+1x2)(x2−1+1x2)dx−2∫x2(x3)2+1dx
I=tan−1(x−1x)−23tan−1(x3)+c
So, f(x)=x−1x and g(x)=x3