Advertisements
Advertisements
Question
`int (3x + 4)/sqrt(2x^2 + 2x + 1) "d"x`
Advertisements
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
APPEARS IN
RELATED QUESTIONS
Evaluate : `int x^2/((x^2+2)(2x^2+1))dx`
Find: `I=intdx/(sinx+sin2x)`
Evaluate: `∫8/((x+2)(x^2+4))dx`
Integrate the rational function:
`(1 - x^2)/(x(1-2x))`
Integrate the rational function:
`x/((x -1)^2 (x+ 2))`
Integrate the rational function:
`1/(x(x^n + 1))` [Hint: multiply numerator and denominator by xn − 1 and put xn = t]
Integrate the rational function:
`(2x)/((x^2 + 1)(x^2 + 3))`
`int (xdx)/((x - 1)(x - 2))` equals:
Find `int(e^x dx)/((e^x - 1)^2 (e^x + 2))`
Find :
`∫ sin(x-a)/sin(x+a)dx`
Integrate the following w.r.t. x : `(2x)/(4 - 3x - x^2)`
Integrate the following w.r.t. x : `(x^2 + x - 1)/(x^2 + x - 6)`
Integrate the following w.r.t. x : `(12x^2 - 2x - 9)/((4x^2 - 1)(x + 3)`
Integrate the following w.r.t. x : `(2x)/((2 + x^2)(3 + x^2)`
Integrate the following w.r.t. x : `2^x/(4^x - 3 * 2^x - 4`
Integrate the following w.r.t. x : `(1)/(x(1 + 4x^3 + 3x^6)`
Integrate the following w.r.t. x: `(1)/(sinx + sin2x)`
Integrate the following w.r.t. x : `(1)/(sin2x + cosx)`
Integrate the following w.r.t. x : `(2log x + 3)/(x(3 log x + 2)[(logx)^2 + 1]`
Integrate the following w.r.t. x: `(x^2 + 3)/((x^2 - 1)(x^2 - 2)`
Integrate the following w.r.t.x : `(1)/(sinx + sin2x)`
Evaluate: `int "3x - 2"/(("x + 1")^2("x + 3"))` dx
Evaluate: `int 1/("x"("x"^5 + 1))` dx
`int "e"^(3logx) (x^4 + 1)^(-1) "d"x`
`int (sinx)/(sin3x) "d"x`
`int sin(logx) "d"x`
`int x^3tan^(-1)x "d"x`
`int x sin2x cos5x "d"x`
Evaluate:
`int (5e^x)/((e^x + 1)(e^(2x) + 9)) dx`
`int xcos^3x "d"x`
`int (sin2x)/(3sin^4x - 4sin^2x + 1) "d"x`
`int (5(x^6 + 1))/(x^2 + 1) "d"x` = x5 – ______ x3 + 5x + c
`int 1/x^3 [log x^x]^2 "d"x` = p(log x)3 + c Then p = ______
Evaluate `int x^2"e"^(4x) "d"x`
Evaluate the following:
`int (x^2 "d"x)/((x^2 + "a"^2)(x^2 + "b"^2))`
Evaluate the following:
`int (2x - 1)/((x - 1)(x + 2)(x - 3)) "d"x`
Evaluate the following:
`int sqrt(tanx) "d"x` (Hint: Put tanx = t2)
If `int "dx"/((x + 2)(x^2 + 1)) = "a"log|1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`, then ______.
The numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1, the denominator becomes eight times the numerator. Find the fraction.
Find : `int (2x^2 + 3)/(x^2(x^2 + 9))dx; x ≠ 0`.
Evaluate`int(5x^2-6x+3)/(2x-3)dx`
Evaluate:
`int 2/((1 - x)(1 + x^2))dx`
