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Question
Evaluate the following.
`int "x"^3/sqrt(1 + "x"^4)` dx
Solution
Let I = `int "x"^3/sqrt(1 + "x"^4)` dx
Put 1 + x4 = t
∴ 4x3 . dx = dt
∴ x3 . dx = `1/4` dt
∴ I = `1/4 int "dt"/sqrt"t"`
`= 1/4 int "t"^((-1)/2)`dt
`= 1/4 * "t"^(1/2)/(1/2)` + c
`= 1/2 sqrt"t" + "c"`
∴ I = `1/2 sqrt(1 + "x"^4)` + c
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