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प्रश्न
Find the integrals of the following:
`1/((x + 1)^2 - 25)`
उत्तर
`int ("d"x)/((x + 1)^2 - 25)`
Put x + 1 = t
dx = dt
`int "dt"/((x + 1)^2 - 5^2) = int "dt"/("t"^2 - 5^2)`
`int ("d"x)/(x^2 - "a"^2) = 1/(2"a") log |(x - "a")/(x + "a")| + "c"`
= `1/(2 xx 5) log |("t" - 5)/("t" + 5)| + "c"`
= `1/10 log |("t" - 5)/("t" + 5)| + "c"`
= `1/10 log |(x + 1 - 5)/(x + 1 + 5)| + "c"`
`int ("d"x)/((x + 1)^2 - 25) = 1/10 log |(x - 4)/(x + 6)| + "c"`
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