Advertisements
Advertisements
Question
Integrate the following functions w.r.t. x : `sqrt(2x^2 + 3x + 4)`
Advertisements
Solution
Let I = `int sqrt(2x^2 + 3x + 4).dx`
= `sqrt(2) int sqrt(x^2 + 3/2 x + 2).dx`
= `sqrt(2) int sqrt((x^2 + 3/2x + 9/16) - 9/16 + 2).dx`
= `sqrt(2) int sqrt((x + 3/4)^2 + (sqrt(23)/4)^2).dx`
= `sqrt(2)[((x + 3/4))/(2) sqrt((x + 3/4)^2 + (sqrt(23)/4)^2 ) + ((23/16))/(2)log|(x + 3/4) + sqrt((x + 3/4)^2 + (sqrt(23)/4)^2)|] + c`
= `ssqrt(2)[((4x + 3)/8) sqrt(x^2 + 3/2x + 2) + (23)/(32)log|(x + 3/4) + sqrt(x^2 + 3/2x + 2)|] + c`.
APPEARS IN
RELATED QUESTIONS
If `int_(-pi/2)^(pi/2)sin^4x/(sin^4x+cos^4x)dx`, then the value of I is:
(A) 0
(B) π
(C) π/2
(D) π/4
If u and v are two functions of x then prove that
`intuvdx=uintvdx-int[du/dxintvdx]dx`
Hence evaluate, `int xe^xdx`
Integrate the function in tan-1 x.
Integrate the function in `(xe^x)/(1+x)^2`.
`int e^x sec x (1 + tan x) dx` equals:
Evaluate the following : `int x^2.log x.dx`
Evaluate the following:
`int sec^3x.dx`
Evaluate the following : `int x.sin^2x.dx`
Evaluate the following : `int (t.sin^-1 t)/sqrt(1 - t^2).dt`
Integrate the following functions w.r.t. x : `e^(2x).sin3x`
Integrate the following functions w.r.t.x:
`e^-x cos2x`
Integrate the following functions w.r.t. x : `sec^2x.sqrt(tan^2x + tan x - 7)`
Integrate the following functions w.r.t. x: `sqrt(x^2 + 2x + 5)`.
Integrate the following functions w.r.t. x : `e^x/x [x (logx)^2 + 2 (logx)]`
Integrate the following functions w.r.t. x : `e^(sin^-1x)*[(x + sqrt(1 - x^2))/sqrt(1 - x^2)]`
If f(x) = `sin^-1x/sqrt(1 - x^2), "g"(x) = e^(sin^-1x)`, then `int f(x)*"g"(x)*dx` = ______.
Choose the correct options from the given alternatives :
`int cos -(3)/(7)x*sin -(11)/(7)x*dx` =
Integrate the following with respect to the respective variable : `t^3/(t + 1)^2`
Integrate the following w.r.t.x : e2x sin x cos x
Evaluate the following.
`int x^2 e^4x`dx
Evaluate the following.
`int x^2 *e^(3x)`dx
Choose the correct alternative from the following.
`int (("x"^3 + 3"x"^2 + 3"x" + 1))/("x + 1")^5 "dx"` =
Evaluate: `int "dx"/("9x"^2 - 25)`
Evaluate: `int "dx"/(25"x" - "x"(log "x")^2)`
Evaluate:
∫ (log x)2 dx
`int sqrt(tanx) + sqrt(cotx) "d"x`
`int (x^2 + x - 6)/((x - 2)(x - 1)) "d"x` = x + ______ + c
`int 1/sqrt(x^2 - 8x - 20) "d"x`
∫ log x · (log x + 2) dx = ?
`int log x * [log ("e"x)]^-2` dx = ?
`int "e"^x [x (log x)^2 + 2 log x] "dx"` = ______.
`int "e"^x int [(2 - sin 2x)/(1 - cos 2x)]`dx = ______.
State whether the following statement is true or false.
If `int (4e^x - 25)/(2e^x - 5)` dx = Ax – 3 log |2ex – 5| + c, where c is the constant of integration, then A = 5.
`int(logx)^2dx` equals ______.
`int_0^1 x tan^-1 x dx` = ______.
Find: `int e^(x^2) (x^5 + 2x^3)dx`.
`intsqrt(1+x) dx` = ______
Solution of the equation `xdy/dx=y log y` is ______
Solve the differential equation (x2 + y2) dx - 2xy dy = 0 by completing the following activity.
Solution: (x2 + y2) dx - 2xy dy = 0
∴ `dy/dx=(x^2+y^2)/(2xy)` ...(1)
Puty = vx
∴ `dy/dx=square`
∴ equation (1) becomes
`x(dv)/dx = square`
∴ `square dv = dx/x`
On integrating, we get
`int(2v)/(1-v^2) dv =intdx/x`
∴ `-log|1-v^2|=log|x|+c_1`
∴ `log|x| + log|1-v^2|=logc ...["where" - c_1 = log c]`
∴ x(1 - v2) = c
By putting the value of v, the general solution of the D.E. is `square`= cx
`inte^(xloga).e^x dx` is ______
Evaluate:
`int e^(ax)*cos(bx + c)dx`
Evaluate the following.
`intx^3 e^(x^2) dx`
Complete the following activity:
`int_0^2 dx/(4 + x - x^2) `
= `int_0^2 dx/(-x^2 + square + square)`
= `int_0^2 dx/(-x^2 + x + 1/4 - square + 4)`
= `int_0^2 dx/ ((x- 1/2)^2 - (square)^2)`
= `1/sqrt17 log((20 + 4sqrt17)/(20 - 4sqrt17))`
Evaluate the following.
`intx^3/sqrt(1+x^4) dx`
Evaluate `int (1 + x + x^2/(2!))dx`
Evaluate:
`inte^x "cosec" x(1 - cot x)dx`
