Sum

**Evaluate the following.**

`int "x"^3/(16"x"^8 - 25)` dx

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#### Solution

Let I = `int "x"^3/(16"x"^8 - 25)` dx

Put x^{4} = t

∴ 4x^{3} dx = dt

∴ x^{3} 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)|` + xc

∴ I = = `1/160 log |(4"x"^4 - 5)/(4"x"^4 + 5)|` + c

#### Notes

[**Note:** Answer in the textbook is incorrect.]

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