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Properties of Inequalities

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Question

$\int \frac{x^{3}}{(x - 1) (x - 2) (x - 3)} d x$

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Solution

$\int \frac{x^{3}}{(x - 1) (x - 2) (x - 3)} d x = \int \frac{x^{3}}{(x - 1) (x^{2} - 5 x + 6)} d x = \int \frac{x^{3}}{x^{3} - 5 x^{2} + 6 x - x^{2} + 5 x - 6} d x = \int \frac{x^{3}}{x^{3} - 6 x^{2} + 11 x - 6} d x ∴ \frac{x^{3}}{x^{3} - 6 x^{2} + 11 x - 6} = 1 + \frac{6 x^{2} + 11 x + 6}{x^{2} - 6 x^{2} + 11 x - 6} \Rightarrow \frac{x^{3}}{x^{3} - 6 x^{2} + 11 x - 6} = 1 + \frac{6 x^{2} - 11 x + 6}{(x - 1) (x - 2) (x - 3)} ∴ \int \frac{x^{3}}{(x - 1) (x - 2) (x - 3)} d x = \int d x + \int \frac{(6 x^{2} - 11 x + 6)}{(x - 1) (x - 2) (x - 3)} d x . . . (1)$

$Let \frac{6 x^{2} - 11 x + 6}{(x - 1) (x - 2) (x - 3)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3} \Rightarrow \frac{6 x^{2} - 11 x + 6}{(x - 1) (x - 2) (x - 3)} = \frac{A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2)}{(x - 1) (x - 2) (x - 3)} \Rightarrow 6 x^{2} - 11 x + 6 = A (x - 2) (x - 3) + B (x - 1) (x - 3) + C (x - 1) (x - 2) . . . (2) Putting x - 2 = 0 or x = 2 in eq (2) \Rightarrow 6 \times 4 - 22 + 6 = B (2 - 1) (2 - 3) \Rightarrow 8 = B (- 1) \Rightarrow B = - 8 Putting x - 3 = 0 or x = 3 in eq (2) \Rightarrow 6 \times 3^{2} - 11 \times 3 + 6 = C (3 - 1) (3 - 2) \Rightarrow 27 = C (2) (1) \Rightarrow C = \frac{27}{2} Putting x - 1 = 0 or x = 1 in eq (2) \Rightarrow 6 \times 1 - 11 + 6 = A (1 - 2) (1 - 3) \Rightarrow 1 = A (- 1) (- 2) \Rightarrow A = \frac{1}{2} ∴ \frac{6 x^{2} - 11 x + 6}{(x - 1) (x - 2) (x - 3)} = \frac{1}{2 (x - 1)} - \frac{8}{x - 2} + \frac{27}{2 (x - 3)} . . . (3) From eq (2) and eq (3) ∴ \int \frac{x^{3} d x}{(x - 1) (x - 2) (x - 3)} = \int d x + \frac{1}{2} \int \frac{1}{x - 1} d x - 8 \int \frac{1}{x - 2} d x + \frac{27}{2} \int \frac{1}{x - 3} d x = x + \frac{1}{2} \ln |x - 1| - 8 \ln |x - 2| + \frac{27}{2} \ln |x - 3| + C$

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Q. The solution of the differential equation

\frac{d y}{d x} + x (x + y) = x^{3} (x + y)^{3} - 1

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\frac{(k + 2)^{3}}{k^{4}} =

I_{1} = \int_{0}^{1} 2^{x^{2}} d x, I_{2} = \int_{0}^{1} 2^{x^{3}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x

I_{4} = \int_{1}^{2} 2^{x^{3}} d x

I_{1} = \int_{0}^{1} 2^{x^{3}} d x, I_{2} = \int_{0}^{1} 2^{x^{2}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x

I_{4} = \int_{1}^{2} 2^{x^{3}} d x

I_{1} = \int_{0}^{1} 2^{x^{2}} d x, I_{2} = \int_{0}^{1} 2^{x^{3}} d x, I_{3} = \int_{1}^{2} 2^{x^{2}} d x

I_{4} = \int_{1}^{2} 2^{x^{3}} d x

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