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Question
Integrate the following with respect to x:
`(2x + 3)/sqrt(x^2 + 4x + 1)`
Solution
Let 2x + 3 = `"A" "d"/("d"x) (x^2 + 4x + 1) + "B"`
2x + 3 = A(2x + 3) + B
2A = 2
⇒ A = 1
4A + B = 3
4 × 1 + B = 3
⇒ B = 3 – 4
B = – 1
(2x + 3) = (2x + 4) – 1
`int (2x + 3)/sqrt(x^2 + 4x + 1) "d"x = int ((2x + 4) - 1)/sqrt(x^2 + 4x + 1) "d"x`
= `int ((2x + 4))/sqrt(x^2 + 4x + 1) "d"x - int ("d"x)/sqrt(x^2 + 4x + 1)`
Put x2 + 4x + 1 = t
(2x + 4) dx = dt
= `intt "dt"/sqrt("t") - int ("d"x)/sqrt((x + 2)^2 - 2^2 + 1)`
= `int "t"^(- 1/2) "dt" - int ("d"x)/sqrt((x + 2)^2 - 3)`
= `("t"^(- 1/2 + 1))/(- 1/2 + 1) - int ("d"x)/sqrt((x + 2)^2 - (sqrt(3))^2`
Put x + 2 = u
dx = dt
= `"t"^(1/2)/(1/2) - int "du"/sqrt("u"^2 - (sqrt(3))^2`
= `2sqrt("t") - log |"u" + sqrt("u"^2 - (sqrt(3))^2)| + "c"`
= `2 sqrt(x^2 + 4x + 1) - log |(x + 2) + sqrt((x + 2)^2 - (sqrt(3))^2)| + "c"`
= `2 sqrt(x^2 + 4x + 1) - log|(x + 2) + sqrt(x^2 + 4x + 4 - 3)| + "c"`
`int (2x + 3)/sqrt(x^2 + 4x + 1) "d"x = 2 sqrt(x^2 + 4x + 1) - log|(x + 2) + sqrt(x^2 + 4x + 1)| + "c"`
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