English

Solve: ∫4x2+5dx

Advertisements
Advertisements

Question

Solve: `int sqrt(4x^2 + 5)dx`

Sum
Advertisements

Solution

`int sqrt(4x^2 + 5).dx = int sqrt((x^2 + 5/4)).dx`

= `2int sqrt(x^2 + 5/4).dx`

= `2int sqrt(x^2 + (sqrt(5)/2)^2).dx`

= `2[x/2 sqrt(x^2 + 5/4) + (5/4)/2 log|x + sqrt(x^2 + 5/4)|] + c_1`

∵ `int sqrt(x^2 + a^2).dx = x/2 sqrt(x^2 + a^2) + a^2/2 log|x + sqrt(x^2 + a^2)| + c`

 = `xsqrt(x^2 + 5/4) + 5/4 log|x + sqrt(x^2 + 5/4)| + c_1`

= `x/2 sqrt(4x^2 + 5) + 5/4 log|x + sqrt((4x^2 + 5)/2)| + c_1`

= `x/2 sqrt(4x^2 + 5) + 5/4 log|(2x + sqrt(4x^2 + 5))/2| + c_1`

= `x/2 sqrt(4x^2 + 5) + 5/4 log|2x + sqrt(4x^2 + 5)| - 5/4 log 2 + c`

= `x/2 sqrt(4x^2 + 5) + 5/4 log|2x + sqrt(4x^2 + 5)| + c_1`

Where c = c1 – `5/4` log2, a constant.

shaalaa.com
  Is there an error in this question or solution?
2025-2026 (March) Model set 2 by shaalaa.com

RELATED QUESTIONS

`int1/xlogxdx=...............`

(A)log(log x)+ c

(B) 1/2 (logx )2+c

(C) 2log x + c

(D) log x + c


Integrate the function in `((x- 3)e^x)/(x - 1)^3`.


Evaluate the following : `int log(logx)/x.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 .(1/x - 1/x^2)`


Integrate the following with respect to the respective variable : `t^3/(t + 1)^2`


Integrate the following w.r.t.x : `(1)/(x^3 sqrt(x^2 - 1)`


Evaluate the following.

∫ x log x dx


Evaluate the following.

`int (log "x")/(1 + log "x")^2` dx


Choose the correct alternative from the following.

`int (("e"^"2x" + "e"^"-2x")/"e"^"x") "dx"` = 


`int 1/(4x + 5x^(-11))  "d"x`


`int (sin(x - "a"))/(cos (x + "b"))  "d"x`


`int (cos2x)/(sin^2x cos^2x)  "d"x`


Choose the correct alternative:

`int ("d"x)/((x - 8)(x + 7))` =


`int 1/(x^2 - "a"^2)  "d"x` = ______ + c


Evaluate `int 1/(x log x)  "d"x`


`int cot "x".log [log (sin "x")] "dx"` = ____________.


Find `int_0^1 x(tan^-1x)  "d"x`


Evaluate the following:

`int_0^1 x log(1 + 2x)  "d"x`


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


Evaluate `int(1 + x + (x^2)/(2!))dx`


Evaluate:

`int e^(ax)*cos(bx + c)dx`


If u and v are two differentiable functions of x, then prove that `intu*v*dx = u*intv  dx - int(d/dx u)(intv  dx)dx`. Hence evaluate: `intx cos x  dx`


Evaluate the following.

`intx^3  e^(x^2) dx`


Evaluate the following:

`intx^3e^(x^2)dx` 


The value of `inta^x.e^x dx` equals


The value of `int (x sin^-1)/(sqrt(1 - x^2)) dx` is equal to:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×