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Question
Integrate the following with respect to x:
`(5x - 2)/(2 + 2x + x^2)`
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Solution
Let 5x – 2 = `"A" "d"/("d"x) (x^2 + 2x + 2) + "B"`
5x – 2 = A(2x + 2) + B
5x – 2 = 2Ax + 2A + B
2A = 5
⇒ A = `5/2`
2A + B = – 2
`2 xx 5/2 + "B"` = – 2
⇒ B = – 2 – 5 = – 7
5x – 2 = `- 5/2 (2x + 2) - 7`
`int (5x - 2)/(x^2 + 2x + 2) "d"x = int (5/2 (2x + 2) - 7)/(x^2 + 2x + 2) "d"x`
= `5/2 int (2x + 2)/(x^2 + 2x + 2) "d"x - 7 int ("d"x)/(x^2 + 2x + 2)`
Put x2 + 2x + 12 = t
(2x + 2)dx = dt
= `5/2 int "dt"/"t" - 7 int ("d"x)/((x + 1)^2 - 1^2 + 2)`
= `5/2 log |"t"| - 7 int ("d"x)/((x + 1)^2 + 1)`
= `5/2 log |x^2 + 2x + 2| - 7 int ("d"x)/((x + 1)^2 + 1^2)`
Put x + 1 = u
dx = du
= `5/2 log |x^2 + 2x + 2| - 7 int "du"/("u"^2 + 1^2)`
= `5/2 log |x^2 + 2x + 2| - 7 xx 1/1 tan^-1 ("u"/1) + "c"`
= `5/2 log|x^2 + 2x + 2| - 7 tan^-1 (x + 1) + "c"`
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