Advertisements
Advertisements
प्रश्न
`int (2x - 7)/sqrt(4x- 1) dx`
Advertisements
उत्तर
Let `I = int (2x - 7)/sqrt(4x - 1) dx`
Let 2x − 7 = A(4x − 1) + B
∴ 2x – 7 = 4Ax + (B – A)
By equating the coefficients on both sides, we get
4A = 2 and B – A = –7
∴ A = `1/2` and B = –7 + A
= `-7 + 1/2`
= `-13/2`
∴ `2x - 7 = 1/2(4x - 1) - 13/2`
∴ `I = 1/2 int ((4x - 1) - 13)/sqrt(4x - 1) dx`
= `1/2 int((4x - 1)/sqrt(4x - 1) - 13/sqrt(4x- 1)) dx`
= `1/2 int(sqrt(4x - 1) - 13/sqrt(4x - 1)) dx`
= `1/2 int(4x- 1)^(1/2) "d"x - 13/2 int(4x - 1)^(1/2) dx`
= `1/2[((4x - 1)^(3/2))/(3/2) xx 1/4] - 13/2 [((4x - 1)^(1/2))/(1/2) xx 1/4]`
∴ `I = 1/12(4x - 1)^(3/2) - 13/4 sqrt(4x - 1) +c`
APPEARS IN
संबंधित प्रश्न
Find : `int x^2/(x^4+x^2-2) dx`
Integrate the rational function:
`x/((x + 1)(x+ 2))`
Integrate the rational function:
`x/((x-1)(x- 2)(x - 3))`
Integrate the rational function:
`(1 - x^2)/(x(1-2x))`
Integrate the rational function:
`(3x -1)/(x + 2)^2`
Integrate the rational function:
`1/(e^x -1)`[Hint: Put ex = t]
Evaluate : `∫(x+1)/((x+2)(x+3))dx`
Integrate the following w.r.t. x : `(12x + 3)/(6x^2 + 13x - 63)`
Integrate the following w.r.t. x : `((3sin - 2)*cosx)/(5 - 4sin x - cos^2x)`
Integrate the following w.r.t. x : `(1)/(2sinx + sin2x)`
Integrate the following w.r.t. x : `(1)/(sinx*(3 + 2cosx)`
Choose the correct options from the given alternatives :
If `int tan^3x*sec^3x*dx = (1/m)sec^mx - (1/n)sec^n x + c, "then" (m, n)` =
Integrate the following with respect to the respective variable : `(6x + 5)^(3/2)`
Integrate the following w.r.t. x: `(2x^2 - 1)/(x^4 + 9x^2 + 20)`
Integrate the following w.r.t. x: `(x^2 + 3)/((x^2 - 1)(x^2 - 2)`
Integrate the following with respect to the respective variable : `cot^-1 ((1 + sinx)/cosx)`
Integrate the following w.r.t.x : `x^2/sqrt(1 - x^6)`
Integrate the following w.r.t.x : `(1)/(sinx + sin2x)`
Integrate the following w.r.t.x : `sec^2x sqrt(7 + 2 tan x - tan^2 x)`
Integrate the following w.r.t.x : `sqrt(tanx)/(sinx*cosx)`
Evaluate: `int 1/("x"("x"^5 + 1))` dx
Evaluate: `int (2"x"^3 - 3"x"^2 - 9"x" + 1)/("2x"^2 - "x" - 10)` dx
`int ((x^2 + 2))/(x^2 + 1) "a"^(x + tan^(-1_x)) "d"x`
`int sqrt((9 + x)/(9 - x)) "d"x`
`int x sin2x cos5x "d"x`
`int ("d"x)/(x^3 - 1)`
`int (3"e"^(2x) + 5)/(4"e"^(2x) - 5) "d"x`
Evaluate `int (2"e"^x + 5)/(2"e"^x + 1) "d"x`
Evaluate the following:
`int x^2/(1 - x^4) "d"x` put x2 = t
Evaluate the following:
`int (2x - 1)/((x - 1)(x + 2)(x - 3)) "d"x`
Evaluate the following:
`int "e"^(-3x) cos^3x "d"x`
If `int 1/((x^2 + 4)(x^2 + 9))dx = A tan^-1 x/2 + B tan^-1(x/3) + C`, then A – B = ______.
Evaluate.
`int (5x^2 - 6x + 3) / (2x -3) dx`
Evaluate:
`int (x + 7)/(x^2 + 4x + 7)dx`
Value of ∫ `(x^2 + 1)/((x − 1)(x − 2))`dx is ______.
When is a rational function called improper?
Which decomposition corresponds to \[\frac{\mathrm{p}x^{2}+\mathrm{q}x+\mathrm{r}}{(x-\mathrm{a})(x^{2}+\mathrm{b}x+\mathrm{c})}\]?
What is done after long division, before writing the appropriate partial-fraction decomposition?
How are the constants \[\mathrm{A},\mathrm{B},\mathrm{C},\ldots\] determined in a partial-fraction decomposition?
Which pair of equations is obtained by equating coefficients in \[5x-5=\mathrm{A}(x-3)+\mathrm{B}(x-2)\]?
What are the values of \[\mathrm{A}\] and \[\mathrm{B}\] in \[\frac{5x-5}{(x-2)(x-3)}=\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-3}\]?
Which decomposition is correct for \[\frac{x^{2}+1}{x^{2}-5x+6}\]?
Evaluate \[\int\frac{x^{2}+1}{x^{2}-5x+6}\,dx\].
What numerator should be used for each distinct linear factor in a partial-fraction decomposition?
