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Question

If x4+1x6+1 dx=tan1(f(x))23tan1(g(x))+c, where c is arbitrary constant, then which of the following is/are correct?

A
f(x) is odd function.
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B
g(x) is odd function.
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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+c1
(c1 is a arbitrary constant)
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Solution

The correct option is D f(x)g(x) dx=1x+13x3+c1
(c1 is a arbitrary constant)
x4+1x6+1 dx=tan1(f(x))23tan1(g(x))+c
Let
I=(x2+1)22x2(x2+1)(x4x2+1) dxI=(x2+1)dx(x4x2+1)2x2 dx(x6+1)I=(1+1x2)dx(x21+1x2)2x2 dx(x3)2+1I=(1+1x2)dx1+(x1x)22x2 dx(x3)2+1
Assuming I1=(1+1x2)dx1+(x1x)2
Let (x1x)=t
I1=11+t2 dtI1=tan1t=tan1(x1x)
I2=x2 dx(x3)2+1
Let x3=t
I2=13tan1(x3)
So,
I=tan1(x1x)23tan1(x3)+cf(x)=x1x, g(x)=x3
f(x) and g(x) are odd functions.

When f(x)=g(x), then
x1x=x3x4x2+1=0(x212)2+34=0
So, no real roots.

Also,
f(x)g(x) dx=x1xx3 dxf(x)g(x) dx=(1x21x4)dxf(x)g(x) dx=1x+13x3+c1

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