# ∫ X 2 + 1 X 4 + X 2 + 1 D X - Mathematics

Sum
$\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{ dx }$

#### Solution

$\text{ We have,}$
$I = \int \frac{\left( x^2 + 1 \right)dx}{x^4 + x^ 2 + 1}$
$\text{Dividing numerator and denominator by x^2 , we get}$
$I = \int \frac{\left( 1 + \frac{1}{x^2} \right)dx}{x^2 + 1 + \frac{1}{x^2}}$
$= \int \frac{\left( 1 + \frac{1}{x^2} \right)dx}{x^2 + \frac{1}{x^2} - 2 + 3}$
$= \int \frac{\left( 1 + \frac{1}{x^2} \right)dx}{\left( x - \frac{1}{x} \right)^2 + 3}$
$\text{ Putting x }- \frac{1}{x} = t$
$\Rightarrow \left( 1 + \frac{1}{x^2} \right)dx = dt$
$\therefore I = \int \frac{dt}{t^2 + 3}$
$= \int\frac{dt}{t^2 + \left( \sqrt{3} \right)^2}$
$= \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{t}{\sqrt{3}} \right) + C$
$= \frac{1}{\sqrt{3}} \tan^{- 1} \left[ \frac{x - \frac{1}{x}}{\sqrt{3}} \right] + C$
$= \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{x^2 - 1}{\sqrt{3} x} \right) + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 1 | Page 190