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प्रश्न
Integrate the following with respect to x.
`1/(2x^2 - 9)`
उत्तर
`int 1/(2x^2 - 9) "d"x = int 1/((sqrt(2x))^2 - 3^2) "d"x`
Let t = `sqrt(2)x`
Then dt = `sqrt(2) "dt"`
= `1/2 int "dt"/("t"^2 - 3^2)`
= `1/sqrt(2) 1/(2(3)) log |("t" - 3)/("t" + 3)| + "c"`
= `1/(6sqrt(2)) log |(sqrt(2x) - 3)/(sqrt(2x) + 3)| + "c"`
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