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प्रश्न
Integrate the following with respect to x:
`(x + 2)/sqrt(x^2 - 1)`
उत्तर
Let x + 2 = `"A" "d"/("d"x) (x^2 - 1) + "B"`
x + 2 = A(2x) + B
2A = 1
⇒ A = `1/2`
B = 2
x + 2 = `1/2 (2x) + 2`
`int (x + 2)/sqrt(x^2 - 1) "d"x = int (1/2 (2x) + 2)/sqrt(x^2 - 1) "d"x`
= `1/2 int (2x)/sqrt(x^2 - 1) "d"x + 2 int ("d"x)/sqrt(x^2 - 1)`
Put `x^2 - 1` = t
2x dx = dt
= `1/2 int "dt"/"t" + 2 int ("d"x)/sqrt(x^2 - 1^2)`
= `1/2 "t"^(- 1/2) "dt" + 2 log |x + sqrt(x^2 - 1^2)| + "c"`
= `1/2 ("t"^(- 1/2 + 1))/(- 1/2 + 1) + 2 log |x + sqrt(x^2 - 1)| + "c"`
= `1/2 ("t"^(1/2))/(1/2) + 2 log |x + sqrt(x^2 - 1)| + "c"`
`int (x + 2)/sqrt(x^2 - 1) "d"x = sqrt(x^2 - 1) + 2 log |x + sqrt(x^2 - 1)| + "c"`
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