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Integrate the following w.r.t. x : 5x2+20x+6x3+2x2+x - Mathematics and Statistics

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Question

Integrate the following w.r.t. x : $\frac{5 x^{2} + 20 x + 6}{x^{3} + 2 x^{2} + x}$

Sum

Solution

Let I = $\int \frac{5 x^{2} + 20 x + 6}{x^{3} + 2 x^{2} + x} \cdot d x$

= $\int \frac{5 x^{2} + 20 x + 6}{x (x^{2} + 2 x + 1)} \cdot d x$

= $\int \frac{5 x^{2} + 20 x + 6}{x {(x + 1)}^{2}} \cdot d x$

Let $\frac{5 x^{2} + 20 x + 6}{x {(x + 1)}^{2}} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{{(x + 1)}^{2}}$

∴ 5x² + 20x + 6 = A(x + 1)² + Bx(x + 1) + Cx
Put x = 0, we get
0 + 0 + 6 = A(1) + B(0)(1) + C(0)
∴ A = 6
Put x + 1 = 0, i x = – 1, we get
5(1) + 20(– 1) + 6 = A(0) + B(– 1)(0) + C(– 1)
∴ – 9 = – C
∴ C = 9
Put x = 1, we get
5(1) + 20(1) + 6 = A(4) + B(1)(2) + C(1)
But A = 6 and C = 9
∴ 31 = 24 + 2B + 9
∴ B = – 1

∴ $\frac{5 x^{2} + 20 x + 6}{x {(x + 1)}^{2}} = \frac{6}{x} - \frac{1}{x + 1} + \frac{9}{{(x + 1)}^{2}}$

∴ I = $\int [\frac{6}{x} - \frac{1}{x + 1} + \frac{9}{{(x + 1)}^{2}}] \cdot d x$

= $6 \int \frac{1}{x} \cdot d x - \int \frac{1}{x + 1} \cdot d x + 9 \int {(x + 1)}^{- 2} \cdot d x$

= $6 \log | x | - \log | x + 1 | + 9 \cdot \frac{{(x + 1)}^{- 1}}{- 1} + c$

= $\log | x^{6} | - \log | x + 1 | - \frac{9}{(x + 1)} + c$

= $\log | \frac{x^{6}}{x + 1} | - \frac{9}{(x + 1)} + c$ .

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Methods of Integration: Integration Using Partial Fractions

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Q 1.14 Q 1.13 Q 1.15

Chapter 3: Indefinite Integration - Exercise 3.4 [Page 145]

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Balbharati Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board

Chapter 3 Indefinite Integration
Exercise 3.4 | Q 1.14 | Page 145

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