English

Integrate the function in x tan-1 x.

Advertisements
Advertisements

Question

Integrate the function in x tan-1 x.

Sum
Advertisements

Solution

Let `I = int x tan^-1 x dx`

`= tan^-1 x int x  dx - int [(d/dx(tan^-1 x)) int (x  dx)]  dx`

`= tan^-1 x (x^2/2) - int 1/ (1 + x^2) * x^2/2 dx`

`= x^2/2 tan^-1 x - 1/2 int x^2/ (x^2 + 1) dx`

`= x^2/2 tan^-1 x - 1/2 int (x^2 + 1 - 1)/ (1 + x^2)  dx`

`= x^2/2 tan^-1 x - 1/2 int (1 - 1/(1 + x^2)) dx`

`= x^2/2 tan^-1 x - 1/2 (x - tan^-1 x) + C`

`= x^2/2 tan^-1 x - 1/2 x + 1/2 tan^-1 x + C`

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Integrals - Exercise 7.6 [Page 327]

APPEARS IN

NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 7 Integrals
Exercise 7.6 | Q 8 | Page 327

RELATED QUESTIONS

Prove that: `int sqrt(a^2 - x^2) * dx = x/2 * sqrt(a^2 - x^2) + a^2/2 * sin^-1(x/a) + c`


Evaluate the following : `int x^2*cos^-1 x*dx`


Evaluate the following : `int (t.sin^-1 t)/sqrt(1 - t^2).dt`


Evaluate the following : `int x.cos^3x.dx`


Evaluate the following : `int(sin(logx)^2)/x.log.x.dx`


Integrate the following functions w.r.t. x : `e^(2x).sin3x`


Integrate the following functions w.r.t. x : `sqrt((x - 3)(7 - x)`


Integrate the following functions w.r.t. x : [2 + cot x – cosec2x]e 


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 with respect to the respective variable : `(3 - 2sinx)/(cos^2x)`


Integrate the following w.r.t.x : cot–1 (1 – x + x2)


Integrate the following w.r.t.x : `(1)/(xsin^2(logx)`


Integrate the following w.r.t.x : `sqrt(x)sec(x^(3/2))*tan(x^(3/2))`


Evaluate the following.

∫ x log x dx


Evaluate the following.

`int "e"^"x" [(log "x")^2 + (2 log "x")/"x"]` dx


Choose the correct alternative from the following.

`int (("x"^3 + 3"x"^2 + 3"x" + 1))/("x + 1")^5  "dx"` = 


Evaluate: `int "dx"/("x"[(log "x")^2 + 4 log "x" - 1])`


Evaluate: `int "dx"/(5 - 16"x"^2)`


`int 1/sqrt(2x^2 - 5)  "d"x`


`int sin4x cos3x  "d"x`


`int "e"^x [x (log x)^2 + 2 log x] "dx"` = ______.


Evaluate the following:

`int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`


Find: `int e^x.sin2xdx`


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


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 ______.


`int1/sqrt(x^2 - a^2) dx` = ______


`intsqrt(1+x)  dx` = ______


`inte^(xloga).e^x dx` is ______


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


Evaluate the following.

`int (x^3)/(sqrt(1 + x^4))dx`


Evaluate:

`int e^(logcosx)dx`


Evaluate:

`int (logx)^2 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/sqrt(1+x^4)`dx


Evaluate the following.

`intx^2e^(4x)dx`


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


Evaluate.

`int(5x^2 - 6x + 3)/(2x - 3)  dx`


Integration by parts is a method of integration based on which rule of differentiation?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×