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प्रश्न
`int 1/(4x^2 - 20x + 17) "d"x`
उत्तर
Let I = `int 1/(4x^2 - 20x + 17) "d"x`
= `int 1/(4(x^2 - 5x + 17/4)) "d"x`
`(1/2 "coefficient of" x)^2 = (1/2 xx (-5))^2`
= `25/4`
∴ I = `1/4 int 1/(x^2 - 5x + 25/4 - 25/4 + 17/4) "d"x`
= `1/4 int 1/((x - 5/2)^2 - 2) "d"x`
= `1/4 int 1/((x - 5/2)^2 - (sqrt(2))^2) "d"x`
= `1/4 * 1/(2sqrt(2)) log |(x - 5/2 - sqrt(2))/(x - 5/2 + sqrt(2))| + "c"`
∴ I = `1/(8sqrt(2)) log|(2x - 5 - 2sqrt(2))/(2x - 5 + 2sqrt(2))| + "c"`
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