∫3x+42x2+2x+1 dx - Mathematics and Statistics

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Sum

`int (3x + 4)/sqrt(2x^2 + 2x + 1)  "d"x`

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Solution

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    

  Is there an error in this question or solution?
Chapter 2.3: Indefinite Integration - Long Answers III

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