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
Evaluate : `int x^2/((x^2+2)(2x^2+1))dx`
Integrate the rational function:
`(3x - 1)/((x - 1)(x - 2)(x - 3))`
Integrate the rational function:
`(2x)/(x^2 + 3x + 2)`
Integrate the rational function:
`(3x + 5)/(x^3 - x^2 - x + 1)`
Integrate the rational function:
`2/((1-x)(1+x^2))`
Integrate the rational function:
`(cos x)/((1-sinx)(2 - sin x))` [Hint: Put sin x = t]
`int (dx)/(x(x^2 + 1))` equals:
Integrate the following w.r.t. x : `(12x + 3)/(6x^2 + 13x - 63)`
Integrate the following w.r.t. x : `(x^2 + x - 1)/(x^2 + x - 6)`
Integrate the following w.r.t. x : `(1)/(x(1 + 4x^3 + 3x^6)`
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 : `(1)/(2cosx + 3sinx)`
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)`
Evaluate: `int (2"x" + 1)/(("x + 1")("x - 2"))` dx
Evaluate: `int ("x"^2 + "x" - 1)/("x"^2 + "x" - 6)` dx
Evaluate: `int (1 + log "x")/("x"(3 + log "x")(2 + 3 log "x"))` dx
`int "e"^(3logx) (x^4 + 1)^(-1) "d"x`
`int sqrt(4^x(4^x + 4)) "d"x`
`int 1/(x(x^3 - 1)) "d"x`
`int sqrt((9 + x)/(9 - x)) "d"x`
`int sec^3x "d"x`
`int "e"^(sin^(-1_x))[(x + sqrt(1 - x^2))/sqrt(1 - x^2)] "d"x`
`int 1/(sinx(3 + 2cosx)) "d"x`
`int ((2logx + 3))/(x(3logx + 2)[(logx)^2 + 1]) "d"x`
Choose the correct alternative:
`int ((x^3 + 3x^2 + 3x + 1))/(x + 1)^5 "d"x` =
State whether the following statement is True or False:
For `int (x - 1)/(x + 1)^3 "e"^x"d"x` = ex f(x) + c, f(x) = (x + 1)2
`int (3"e"^(2"t") + 5)/(4"e"^(2"t") - 5) "dt"`
If `int(sin2x)/(sin5x sin3x)dx = 1/3log|sin 3x| - 1/5log|f(x)| + c`, then f(x) = ______
Evaluate the following:
`int x^2/(1 - x^4) "d"x` put x2 = t
Evaluate the following:
`int (x^2"d"x)/(x^4 - x^2 - 12)`
Evaluate the following:
`int (x^2 "d"x)/((x^2 + "a"^2)(x^2 + "b"^2))`
If `int "dx"/((x + 2)(x^2 + 1)) = "a"log|1 + x^2| + "b" tan^-1x + 1/5 log|x + 2| + "C"`, then ______.
Find: `int x^2/((x^2 + 1)(3x^2 + 4))dx`
Evaluate:
`int(2x^3 - 1)/(x^4 + x)dx`
If \[\int\frac{2x+3}{(x-1)(x^{2}+1)}\mathrm{d}x\] = \[=\log_{e}\left\{(x-1)^{\frac{5}{2}}\left(x^{2}+1\right)^{a}\right\}-\frac{1}{2}\tan^{-1}x+\mathrm{A}\] where A is an arbitrary constant, then the value of a is
Which partial-fraction decomposition is appropriate for \[\frac{\mathrm{p}x+\mathrm{q}}{(x-\mathrm{a})(x-\mathrm{b})}\]?
Which partial form corresponds to \[\frac{\mathrm{p}x^{2}+\mathrm{q}x+\mathrm{r}}{(x-\mathrm{a})(x-\mathrm{b})(x-\mathrm{c})}\]?
What should be checked before beginning partial-fraction decomposition?
What is the result of dividing \[x^{2}+1\] by \[x^{2}-5x+6\]?
For \[\frac{5x-5}{(x-2)(x-3)}=\frac{\mathrm{A}}{x-2}+\frac{\mathrm{B}}{x-3}\], which equation results after clearing denominators?
