Maharashtra State BoardHSC Commerce 12th Board Exam
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Solve the following : ∫12x+3x(x+2)⋅dx - Mathematics and Statistics

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Sum

Solve the following : `int_1^2 (x + 3)/(x (x + 2))*dx`

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Solution

Let I = `int_1^2 (x + 3)/(x (x + 2))*dx`

Let `(x + 3)/(x(x + 2)) = "A"/x + "B"/(x + 2)`    ...(i)

∴ x + 3 = A(x + 2) + Bx   ...(ii)
Putting x = 0 in (ii), we get
3 = A(0 + 2) + B(0)
∴ 3 = 2A
∴ A = `(3)/(2)`
Putting x = – 2 in (ii), we get
– 2 + 3 = A(–2 + 2) + B(– 2)
∴ 1 = – 2B
∴ B = `(1)/(2)`
From (i), we get
`(x + 3)/(x(x + 2)) = (3)/(2)*(1)/x - (1)/(2(x + 2)`

∴ I = `int_1^2[3/(2x) - (1)/(2(x + 2))]*dx`

= `(3)/(2) int_1^2 (1)/x*dx - (1)/(2) int_1^2 (1)/(x + 2)*dx`

= `(3)/(2)[log|x|]_1^2 - (1)/(2)[log|x + 2|]_1^2`

= `(3)/(2)[log |2| - log|1|] - (1)/(2) [log|2 +2| - log|1 + 2|]`

= `(3)/(2)(log 2 - 0) - (1)/(2)(log4 - log3)`

= `(3)/(2) log2 - (1)/(2)(log  4/3)`

= `(1)/(2)(3log2 - log  4/3)`

= `(1)/(2) log(2^3 xx 3/4)`

= `(1)/(2)log((8 xx 3)/4)`

∴ I = `(1)/(2)log6`.

Concept: Fundamental Theorem of Integral Calculus
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 6 Definite Integration
Miscellaneous Exercise 6 | Q 4.02 | Page 150
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