What is {{{{{x}}^4} - 1{{\;}}}}{{{{{x}}² {{{{x}}^4} + {{{x}}² + 1} }}{{dx}} equal to?

Correct Answer is : x2+1x2+1+c {{{{x}}² + {1}{{{{{x}}²}} + 1} + {{c}}x2+x21+1+c

{{{{{{x}}^4} + {{{x}}² + 1}}{{{x}}}} + {{c}}

{{{{x}}^4} + 2 - {1}{{{{{x}}²}}} + {{c}}

{{{{x}}² + {1}{{{{{x}}²}} + 1} + {{c}}

{{{{{{x}}^4} - {{{x}}² + 1}}{{{x}}} + {{c}}}

Correct Answer is : x2+1x2+1+c {{{{x}}² + {1}{{{{{x}}²}} + 1} + {{c}}x2+x21+1+c

Test Index

UPSC NDA 2 2016 Mathematics Solved Paper

Question : 92 of 120

Marks: +1, -0

What is

\smallint \frac{{{{\rm{x}}^4} - 1{\rm{\;}}}}{{{{\rm{x}}^2}\sqrt {{{\rm{x}}^4} + {{\rm{x}}^2} + 1} }}{\rm{dx}}

equal to?

Solution:

Concept:

Rewrite the integrand to match the derivative of

x^2 + \frac{1}{x^2} + 1

, then substitute to simplify.

Explanation:

Start with

I = \int \frac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} \, dx

Factor

x^2

inside the square root:

\sqrt{x^4 + x^2 + 1} = \sqrt{x^2\left(x^2 + 1 + \frac{1}{x^2}\right)} = x\sqrt{x^2 + 1 + \frac{1}{x^2}}

Then

I = \int \frac{x^4 - 1}{x^3 \sqrt{x^2 + 1 + \frac{1}{x^2}}} \, dx

Divide numerator and denominator by

x^3

I = \int \frac{x - \frac{1}{x^3}}{\sqrt{x^2 + 1 + \frac{1}{x^2}}} \, dx

Let

t^2 = x^2 + 1 + \frac{1}{x^2}

. Differentiate:

2t\,dt = (2x - \frac{2}{x^3})\,dx

, so

t\,dt = (x - \frac{1}{x^3})\,dx

Now

I = \int \frac{t\,dt}{t} = \int dt = t + c

Substitute back:

t = \sqrt{x^2 + 1 + \frac{1}{x^2}}

Answer:

\sqrt{x^2 + \frac{1}{x^2} + 1} + c

(Option C).

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