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Question
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
Options
a = `1/3`, b = 1
a = `(-1)/3`, b = 1
a = `(-1)/3`, b = –1
a = `1/3`, b = –1
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Solution
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then a = `1/3`, b = –1.
Explanation:
Let I = `intx^3/sqrt(1 + x^2) "d"x`
Put 1 + x2 = t
⇒ 2x dx = dt
⇒ x dx = `"dt"/2`
∴ I = `1/2 int "t"/sqrt("t") "dt" - 1/2 int 1/sqrt("t") "dt"`
= `1/2 int sqrt("t") "dt" - 1/2 int "t"^((-1)/2) "dt"`
= `1/2 xx 2/3 ("t")^(3/2) - 1/2 * 2sqrt("t") + "C"`
= `1/3(1 + x^2)^(3/2) - sqrt(1 + x^2) + "C"`
But I = `"a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`
Comparing the like terms we get,
∴ a = `1/3` and b = –1.
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