# ∫ X 3 + X 2 + 2 X + 1 X 2 − X + 1 D X - Mathematics

Sum
$\int\frac{x^3 + x^2 + 2x + 1}{x^2 - x + 1}\text{ dx }$

#### Solution

$\text{ Let } I = \int\left( \frac{x^3 + x^2 + 2x + 1}{x^2 - x + 1} \right) dx$

$\text{ Therefore },$
$\frac{x^3 + x^2 + 2x + 1}{x^2 - x + 1} = x + 2 + \frac{3x - 1}{x^2 - x + 1} . . . . . \left( 1 \right)$
$\text{ Let }$
$3x - 1 = A\frac{d}{dx} \left( x^2 - x + 1 \right) + B$
$3x - 1 = A \left( 2x - 1 \right) + B$
$3x - 1 = \left( 2A \right) x + B - A$
$\text{Equating Coefficients of like } terms$
$2A = 3$
$A = \frac{3}{2}$
$B - A = - 1$
$B - \frac{3}{2} = - 1$
$B = \frac{1}{2}$
$\int\left( \frac{x^3 + x^2 + 2x + 1}{x^2 - x + 1} \right) dx = \int\left( x + 2 \right) dx + \int\left( \frac{\frac{3}{2} \left( 2x - 1 \right) + \frac{1}{2}}{x^2 - x + 1} \right) dx$

$= \int\left( x + 2 \right) dx + \frac{3}{2} \int\left( \frac{2x - 1}{x^2 - x + 1} \right) dx + \frac{1}{2}\int\frac{dx}{x^2 - x + 1}$
$= \int\left( x + 2 \right) dx + \frac{3}{2}\int\frac{\left( 2x - 1 \right) dx}{x^2 - x + 1} + \frac{1}{2}\int\frac{dx}{x^2 - x + \frac{1}{4} - \frac{1}{4} + 1}$
$= \int\left( x + 2 \right) dx + \frac{3}{2}\int\frac{\left( 2x - 1 \right) dx}{x^2 - x + 1} + \frac{1}{2}\int\frac{dx}{\left( x - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}$
$= \frac{x^2}{2} + 2x + \frac{3}{2} \text{ log }\left| x^2 - x + 1 \right| + \frac{1}{2} \times \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{x - \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + C$
$= \frac{x^2}{2} + 2x + \frac{3}{2} \text{ log } \left| x^2 - x + 1 \right| + \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{2x - 1}{\sqrt{3}} \right) + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.2 | Q 8 | Page 106

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