English

∫1x2-a2dx = ______.

Advertisements
Advertisements

Question

`int 1/sqrt(x^2 - a^2)dx` = ______.

Fill in the Blanks
Advertisements

Solution

`int 1/sqrt(x^2 - a^2)dx` = `bb(underline(log|x + sqrt(x^2 - a^2)| + c)`.

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

RELATED QUESTIONS

Integrate the function in tan-1 x.


Integrate the function in x (log x)2.


Integrate the function in `e^x (1 + sin x)/(1+cos x)`.


`int e^x sec x (1 +   tan x) dx` equals:


Find : 

`∫(log x)^2 dx`


Evaluate the following : `int e^(2x).cos 3x.dx`


Integrate the following functions w.r.t. x : `sqrt(2x^2 + 3x + 4)`


Choose the correct options from the given alternatives :

`int (1)/(x + x^5)*dx` = f(x) + c, then `int x^4/(x + x^5)*dx` =


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 : `(sin^6θ + cos^6θ)/(sin^2θ*cos^2θ)`


Evaluate the following.

`int x^3 e^(x^2)`dx


`int ["cosec"(logx)][1 - cot(logx)]  "d"x`


`int ("e"^xlog(sin"e"^x))/(tan"e"^x)  "d"x`


Evaluate `int (2x + 1)/((x + 1)(x - 2))  "d"x`


Evaluate the following:

`int_0^1 x log(1 + 2x)  "d"x`


`int x/((x + 2)(x + 3)) dx` = ______ + `int 3/(x + 3) dx`


Evaluate: `int_0^(pi/4) (dx)/(1 + tanx)`


Find: `int (2x)/((x^2 + 1)(x^2 + 2)) dx`


Find the general solution of the differential equation: `e^((dy)/(dx)) = x^2`.


If `int(2e^(5x) + e^(4x) - 4e^(3x) + 4e^(2x) + 2e^x)/((e^(2x) + 4)(e^(2x) - 1)^2)dx = tan^-1(e^x/a) - 1/(b(e^(2x) - 1)) + C`, where C is constant of integration, then value of a + b is equal to ______.


`int_0^1 x tan^-1 x  dx` = ______.


`intsqrt(1+x)  dx` = ______


`int(3x^2)/sqrt(1+x^3) dx = sqrt(1+x^3)+c`


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 ______


The integrating factor of `ylogy.dx/dy+x-logy=0` is ______.


`int logx  dx = x(1+logx)+c`


Evaluate:

`int((1 + sinx)/(1 + cosx))e^x dx`


Prove that `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`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×