English

Integrate the function 3x1+2x4

Advertisements
Advertisements

Question

Integrate the function `(3x)/(1+ 2x^4)`

Sum
Advertisements

Solution

Let `I = int (3x)/ (1 + 2x^4) dx`

Put x2 = t

⇒ 2x dx = dt

⇒ `x  dx = dt/2`

∴ `I = 3/2 int dt/(1 + 2t^2 )`

`= 3/4 int dt/ (1/2 + t^2)`

`= 3/4 int dt/ ((1/sqrt2)^2 + t^2)`           `....[∵ int dx/(a^2+x^2) = 1/a tan^-1  x/a + C]`

`3/4* 1/ (1/sqrt2) tan^-1 (t/(1/sqrt2)) + C`

`3/(2sqrt2) tan^-1 sqrt2t + C`

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Integrals - Exercise 7.4 [Page 315]

APPEARS IN

NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 7 Integrals
Exercise 7.4 | Q 5 | Page 315

RELATED QUESTIONS

Evaluate : ` int x^2/((x^2+4)(x^2+9))dx`


 

find : `int(3x+1)sqrt(4-3x-2x^2)dx`

 

Find:

`int(x^3-1)/(x^3+x)dx`


Integrate the function `1/sqrt(9 - 25x^2)`


Integrate the function `x^2/(1 - x^6)`


Integrate the function `(x - 1)/sqrt(x^2 - 1)`


Integrate the function `1/sqrt(x^2 +2x + 2)`


Integrate the function `1/(9x^2 + 6x + 5)`


Integrate the function `1/sqrt(7 - 6x - x^2)`


Integrate the function `1/sqrt((x -1)(x - 2))`


Integrate the function `1/sqrt(8+3x  - x^2)`


Integrate the function `1/sqrt((x - a)(x - b))`


Integrate the function `(6x + 7)/sqrt((x - 5)(x - 4))`


`int dx/(x^2 + 2x + 2)` equals:


`int dx/sqrt(9x - 4x^2)` equals:


Integrate the function:

`sqrt(1- 4x^2)`


Integrate the function:

`sqrt(x^2 + 4x + 6)`


Integrate the function:

`sqrt(1-4x - x^2)`


Integrate the function:

`sqrt(x^2 + 3x)`


Integrate the function:

`sqrt(1+ x^2/9)`


`int sqrt(1+ x^2)  dx` is equal to ______.


`int sqrt(x^2 - 8x + 7) dx` is equal to ______.


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


\[\int e^{ax} \text{ sin} \left( bx + C \right) dx\]

\[\int e^{2x} \cos \left( 3x + 4 \right) \text{ dx }\]

\[\int e^{2x} \sin x\ dx\]

\[\int\frac{1}{x^3}\text{ sin } \left( \text{ log x }\right) dx\]

\[\int e^{2x} \cos^2 x\ dx\]

\[\int e^{- 2x} \sin x\ dx\]

\[\int x^2 e^{x^3} \cos x^3 dx\]

\[\int\frac{1}{\left( x^2 - 1 \right) \sqrt{x^2 + 1}} \text{ dx }\]

Integration of \[\frac{1}{1 + \left( \log_e x \right)^2}\] with respect to loge x is


\[\int\frac{8x + 13}{\sqrt{4x + 7}} \text{ dx }\]


Find : \[\int\left( 2x + 5 \right)\sqrt{10 - 4x - 3 x^2}dx\] .


If θ f(x) = `int_0^x t sin t  dt` then `f^1(x)` is


Find `int (dx)/sqrt(4x - x^2)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×