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Question
Evaluate the following:
`int sqrt(1 + x^2)/x^4 "d"x`
Solution
Let I = `int sqrt(1 + x^2)/x^4 "d"x`
= `int sqrt(x^2 (1 + 1/x^2))/x^4 "d"x`
= `int (xsqrt(1 + 1/x^2))/x^4 "d"x`
= `int (xsqrt(1 + 1/x^2))/x^4 "d"x`
Put `1 + 1/x^2` = r2
⇒ `(-2)/x^3 "d"x = 2"t" "dt"`
⇒ `- "dx"/x^3` = t dt
∴ I = `- int "t"^2 "dt"`
= `- "t"^3/3 + "C"`
= `- 1/3(1 + 1/x^2)^(3/2) + "C"`
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