By completing the following activity, Evaluate ∫12x+3x(x+2) dx Solution: Let I = ∫12x+3x(x+2) dx Let x+3x(x+2)=Ax+B(x+2) ∴ x + 3 = A(x + 2) + B. x ∴ A = □, B = □ ∴ I = ∫12[( )x+( )(x+2)]dx ∴ I = - Mathematics and Statistics

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Sum

By completing the following activity, Evaluate `int_1^2 (x + 3)/(x(x + 2))  "d"x`

Solution: Let I = `int_1^2 (x + 3)/(x(x + 2))  "d"x`

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

∴ x + 3 = A(x + 2) + B.x

∴ A = `square`, B = `square`

∴ I = `int_1^2[("( )")/x + ("( )")/((x + 2))] "d"x`

∴ I = `[square log x + square log(x + 2)]_1^2`

∴ I = `square`

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Solution

Let I = `int_1^2 (x + 3)/(x(x + 2))  "d"x`

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

∴ x + 3 = A(x + 2) + B.x      ......(i)

Putting x = – 2 in (i), we get

– 2 + 3 = A(0) + B(– 2)

∴ 1 = – 2B

∴ B = `-1/2`

Putting x = 0 in (i), we get

0 + 3 = A(0 + 2) + B(0)

∴ 3 = 2A

∴ A = `3/2`

∴ A = `3/2`, B = `-1/2` 

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

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

= `3/2 (log 2 - log 1) - 1/2 (log 4 - log 3)`

= `3/2 (log 2 - 0) - 1/2 log (4/3)`

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

= `1/2 (log8 - log  4/3)`

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

∴ I = `1/2 log 6`

Concept: Fundamental Theorem of Integral Calculus
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Chapter 1.6: Definite Integration - Q.6

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