Question

# If ∫x4+1x6+1 dx=tan−1(f(x))−23tan−1(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=tan−1(f(x))−23tan−1(g(x))+c Let I=∫(x2+1)2−2x2(x2+1)(x4−x2+1) dx⇒I=∫(x2+1)dx(x4−x2+1)−2∫x2 dx(x6+1)⇒I=∫(1+1x2)dx(x2−1+1x2)−2∫x2 dx(x3)2+1⇒I=∫(1+1x2)dx1+(x−1x)2−2∫x2 dx(x3)2+1 Assuming I1=∫(1+1x2)dx1+(x−1x)2 Let (x−1x)=t ⇒I1=∫11+t2 dt⇒I1=tan−1t=tan−1(x−1x) I2=∫x2 dx(x3)2+1 Let x3=t ⇒I2=13tan−1(x3) So, I=tan−1(x−1x)−23tan−1(x3)+cf(x)=x−1x, g(x)=x3 f(x) and g(x) are odd functions. When f(x)=g(x), then x−1x=x3⇒x4−x2+1=0⇒(x2−12)2+34=0 So, no real roots. Also, ∫f(x)g(x) dx=∫x−1xx3 dx⇒∫f(x)g(x) dx=∫(1x2−1x4)dx⇒∫f(x)g(x) dx=−1x+13x3+c1

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