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Question
Evaluate the following.
`int "x"^3/(16"x"^8 - 25)` dx
Solution
Let I = `int "x"^3/(16"x"^8 - 25)` dx
Put x4 = t
∴ 4x3 dx = dt
∴ x3 dx = `1/4` dt
∴ I = `1/4 int "dt"/(16"t"^2 - 25)`
`= 1/(4 xx 16) int "dt"/("t"^2 - 25/16)`
`= 1/64 int "dt"/("t"^2 - (5/4)^2)`
`= 1/64 xx 1/(2 xx 5/4) log |("t" - 5/4)/("t" + 5/4)|` + c
`= 1/160 log |("4t" - 5)/("4t" + 5)|` + c
∴ I = = `1/160 log |(4"x"^4 - 5)/(4"x"^4 + 5)|` + c
Notes
The answer in the textbook is incorrect.
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