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Question
If `int "dx"/((x + 2)(x^2 + 1)) = "a"log|1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`, then ______.
Options
a = `(-1)/10`, b = `(-2)/5`
a = `1/10`, b = `- 2/5`
a = `(-1)/10`, b = `2/5`
a = `1/10`, b = `2/5`
Solution
If `int "dx"/((x + 2)(x^2 + 1)) = "a"log|1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`, then a = `(-1)/10`, b = `2/5`.
Explanation:
Given that, `int "dx"/((x + 2)(x^2 + 1)) = "a"log|1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`
Now, I = `int "dx"/((x + 2)(x^2 + 1))`
`1/((x + 2)(x^2 + 1)) = "A"/(x + 2) + ("B"x + "C")/(x^2 + 1)`
⇒ 1 = A(x2 + 1) + (Bx + C)(x + 2)
⇒ 1 = (A + B)x2 + (2B + C)x + A + 2C
Comapring coefficient, we get
A + B = 0
A + 2C = 1
2B + C = 0
Solving we get A = `1/5`
B = `- 1/5`
And C = `2/5`
∴ `int "dx"/((x + 2)(x^2 + 1))`
= `1/5 int 1/(x + 2) "d"x + int (- 1/5 + 2/5)/(x^2 + 1) "d"x`
= `1/5 int 1/(x + 2) "d"x - 1/10 int (2x)/(1 + x^2) "d"x + 1/5 int 2/(1 + x^2) "d"x`
= `1/5 log|x + 2| - 1/10 log|1 + x^2| + 2/5 tan^-1x + "C"`
= `"a" log |1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"` ....(Given)
∴ a = `(-1)/10`, b = `2/5`.
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