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प्रश्न
`int sqrt(("e"^(3x) - "e"^(2x))/("e"^x + 1)) "d"x`
उत्तर
Let I = `int sqrt(("e"^(3x) - "e"^(2x))/("e"^x + 1)) "d"x`
= `int sqrt(("e"^(2x)("e"^x - 1))/("e"^x + 1)) "d"x`
= `int"e"^x sqrt(("e"^x - 1)/("e"^x + 1)) "d"x`
Put ex = t
∴ ex dx = dt
∴ I = `int sqrt(("t" - 1)/("t" + 1)) "dt"`
= `int sqrt(("t" - 1)/("t" + 1) xx ("t" - 1)/("t" - 1)) "dt"`
= `int ("t" - 1)/sqrt("t"^2 - 1) "dt"`
= `int ("t"/sqrt("t"^2 - 1) - 1/sqrt("t"^2 - 1)) "dt"`
= `int "t"/sqrt("t"^2 - 1) "dt" - int 1/sqrt("t"^2 - 1) "dt"`
= I1 − I2 .......(i)
I1 = `int "t"/sqrt("t"^2 - 1) "dt"`
Put t2 − 1 = a
∴ 2t dt = da
∴ I1 = `1/2 int "da"/sqrt("a")`
= `1/2 int "a"^(1/2) "da"`
= `1/2("a"^(1/2)/(1/2)) + "c"_1`
= `sqrt("a") + "c"_1`
= `sqrt("t"^2 - 1) + "c"_1`
∴ I1 = `sqrt("e"^(2x) - 1) + "c"_1` ......(ii)
I2 = `int 1/sqrt("t"^2 - 1^2) "dt"`
= `log|"t" + sqrt("t"^2 - 1^2)| + "c"_2`
∴ I2 = `log|"e"^x + sqrt("e"^(2x) - 1)| + "c"_2` .......(iiii)
From (i), (ii) and (iii), we get
I = `sqrt("e"^(2x) - 1) - log|"e"^x + sqrt("e"^(2x) - 1)| +"c"`,
where c = c1 − c2
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