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