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प्रश्न
`int (3x + 4)/sqrt(2x^2 + 2x + 1) "d"x`
उत्तर
Let I = `int (3x + 4)/sqrt(2x^2 + 2x + 1) "d"x`
Let 3x + 4 = `"A" "d"/("d"x)(2x^2 + 2x + 1) + "B"`
∴ 3x + 4 = A(4x + 2) + B
∴ 3x + 4 = 4Ax + 2A + B
By equating the coefficients on both sides, we get
4A = 3 and 2A + B = 4
∴ A = `3/4` and `2(3/4) + "B"` = 4
∴ B = `5/2`
∴ 3x + 4 = `3/4(4x + 2) + 5/2`
∴ I = `int (3/4(4x + 2) + 5/2)/sqrt(2x^2 + 2x + 1) "d"x`
= `3/4 int (4x + 2)/sqrt(2x^2 + 2x + 1) "d"x + 5/2 int 1/sqrt(2x^2 + 2x + 1) "d"x`
= I1 + I2 ........(i)
I1 = `3/4 int (4x + 2)/sqrt(2x^2 + 2x + 1) "d"x`
Put 2x2 + 2x + 1 = t
∴ (4x + 2) dx = dt
∴ I1 = `3/4 int "dt"/sqrt("t")`
= `3/4 int "t"^(1/2) "dt"`
= `3/4("t"^(1/2)/(1/2)) + "c"_1`
= `3/2 sqrt("t") + "c"_1`
∴ I1 = `3/2 sqrt(2x^2 + 2x + 1) + "c"_1` .........(ii)
I2 = `5/2 int 1/sqrt(2x^2 + 2x + 1) "d"x`
= `5/2 int 1/sqrt(2(x^2 + x + 1/2)) "d"x`
`(1/2 "coefficient of" x)^2 = (1/2 xx 1)^2`
= `1/4`
∴ I2 = `5/(2sqrt(2)) int 1/sqrt(x^2 + x + 1/4 - 1/4 + 1/2) "d"x`
= `5/(2sqrt(2)) int 1/sqrt((x + 1/2)^2 - (1/2)^2) "d"x`
= `5/(2sqrt(2)) log|x + 1/2 + sqrt((x + 1/2)^2 - (1/2)^2)| + "c"_2`
∴ I2 = `5/(2sqrt(2)) log|x + 1/2 + sqrt(x^2 + x + 1/2)| + "c"_2` ........(iii)
From (i), (ii) and (iii), we get
I = `3/2 sqrt(2x^2 + 2x + 1) + 5/(2sqrt(2)) log|x + 1/2 + sqrt(x^2 + x + 1/2)| + "c"`,
where c = c1 + c2
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