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∫ 3 X + 5 X 3 − X 2 − X + 1 D X - Mathematics

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Sum
\[\int\frac{3x + 5}{x^3 - x^2 - x + 1} dx\]
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Solution

We have,

\[I = \int\frac{\left( 3x + 5 \right)dx}{x^3 - x^2 - x + 1}\]

\[ = \int\frac{\left( 3x + 5 \right)dx}{x^2 \left( x - 1 \right) - 1\left( x - 1 \right)}\]

\[ = \int\frac{\left( 3x + 5 \right)dx}{\left( x^2 - 1 \right) \left( x - 1 \right)}\]

\[ = \int\frac{\left( 3x + 5 \right)dx}{\left( x - 1 \right) \left( x + 1 \right) \left( x - 1 \right)}\]

\[ = \int\frac{\left( 3x + 5 \right)dx}{\left( x - 1 \right)^2 \left( x + 1 \right)}\]

\[\text{Let }\frac{3x + 5}{\left( x - 1 \right)^2 \left( x + 1 \right)} = \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{\left( x - 1 \right)^2}\]

\[ \Rightarrow \frac{3x + 5}{\left( x - 1 \right)^2 \left( x + 1 \right)} = \frac{A \left( x - 1 \right)^2 + B\left( x + 1 \right) \left( x - 1 \right) + C\left( x + 1 \right)}{\left( x + 1 \right) \left( x - 1 \right)^2}\]

\[ \Rightarrow 3x + 5 = A\left( x^2 - 2x + 1 \right) + B\left( x^2 - 1 \right) + Cx + C\]

\[ \Rightarrow 3x + 5 = \left( A + B \right) x^2 + \left( - 2A + C \right)x + \left( A - B + C \right)\]

\[\text{Equating coefficient of like terms}\]

\[A + B = 0 . . . . . \left( 1 \right)\]

\[ - 2A + C = 3 . . . . . \left( 2 \right)\]

\[A - B + C = 5 . . . . . \left( 3 \right)\]

\[\text{Solving these three equations, we get}\]

\[A = \frac{1}{2}\]

\[B = - \frac{1}{2}\]

\[C = 4\]

\[ \therefore \frac{3x + 5}{\left( x - 1 \right)^2 \left( x + 1 \right)} = \frac{1}{2\left( x + 1 \right)} - \frac{1}{2\left( x - 1 \right)} + \frac{4}{\left( x - 1 \right)^2}\]

\[ \Rightarrow I = \frac{1}{2}\int\frac{dx}{x + 1} - \frac{1}{2}\int\frac{dx}{x - 1} + 4\int \left( x - 1 \right)^{- 2} dx\]

\[ = \frac{1}{2}\log \left| x + 1 \right| - \frac{1}{2}\log \left| x - 1 \right| - \frac{4}{\left( x - 1 \right)} + C'\]

\[ = \frac{1}{2}\log \left| \frac{x + 1}{x - 1} \right| - \frac{4}{x - 1} + C'\]

Concept: Indefinite Integral Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Q 43 | Page 177

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