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प्रश्न
Evaluate: `int "dx"/(5 - 16"x"^2)`
उत्तर
Let I = `int "dx"/(5 - 16"x"^2)`
`= int 1/(16(5/16 - "x"^2))` dx
`= 1/16 int 1/((sqrt5/4)^2 - "x"^2)` dx
`= 1/16 * 1/(2 sqrt5/4) log |(sqrt5/4 + "x")/(sqrt5/4 - "x")|` + c
∴ I = `1/(8sqrt5) log |(sqrt5 + 4"x")/(sqrt5 - 4"x")|` + c
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