English

∫e3x-e2xex+1 dx

Advertisements
Advertisements

Question

`int sqrt(("e"^(3x) - "e"^(2x))/("e"^x + 1))  "d"x`

Sum
Advertisements

Solution

Let I = `int sqrt(("e"^(3x) - "e"^(2x))/("e"^x + 1))  "d"x`

= `int sqrt(("e"^(2x)("e"^x - 1))/("e"^x + 1))  "d"x`

= `int"e"^x sqrt(("e"^x - 1)/("e"^x + 1))  "d"x`

Put ex = t

∴ ex dx = dt

∴ I = `int sqrt(("t" - 1)/("t" + 1))  "dt"`

= `int sqrt(("t" - 1)/("t" + 1) xx ("t" - 1)/("t" - 1))  "dt"`

= `int ("t" - 1)/sqrt("t"^2 - 1)  "dt"`

= `int ("t"/sqrt("t"^2 - 1) - 1/sqrt("t"^2 - 1))  "dt"`

= `int "t"/sqrt("t"^2 - 1)  "dt" - int  1/sqrt("t"^2 - 1)  "dt"`

= I1 − I2      .......(i)

I1 = `int "t"/sqrt("t"^2 - 1)  "dt"`

Put t2 − 1 = a

∴ 2t dt = da

I1 = `1/2 int "da"/sqrt("a")`

= `1/2 int "a"^(1/2) "da"`

= `1/2("a"^(1/2)/(1/2)) + "c"_1`

= `sqrt("a") + "c"_1`

= `sqrt("t"^2 - 1) + "c"_1`

I1 = `sqrt("e"^(2x) - 1) + "c"_1`    ......(ii)

I2 = `int 1/sqrt("t"^2 - 1^2)  "dt"`

= `log|"t" + sqrt("t"^2 - 1^2)| + "c"_2`

I2 = `log|"e"^x + sqrt("e"^(2x) - 1)| + "c"_2`  .......(iiii)

 From (i), (ii) and (iii), we get

I = `sqrt("e"^(2x) - 1) - log|"e"^x + sqrt("e"^(2x) - 1)| +"c"`,

 where c = c1 − c2

shaalaa.com
  Is there an error in this question or solution?
Chapter 2.3: Indefinite Integration - Long Answers III

RELATED QUESTIONS

Evaluate : `int (sinx)/sqrt(36-cos^2x)dx`


Evaluate :`intxlogxdx`


Find: `int(x+3)sqrt(3-4x-x^2dx)`


Evaluate: `int sqrt(tanx)/(sinxcosx) dx`


Integrate the functions:

`(sin^(-1) x)/(sqrt(1-x^2))`


`(10x^9 + 10^x log_e 10)/(x^10 + 10^x)  dx` equals:


`int (dx)/(sin^2 x cos^2 x)` equals:


 Write a valoue of \[\int \sin^3 x \cos x\ dx\]

 


Write a value of

\[\int e^{2 x^2 + \ln x} \text{ dx}\]

Write a value of\[\int\left( e^{x \log_e \text{  a}} + e^{a \log_e x} \right) dx\] .


Write a value of \[\int\frac{1 - \sin x}{\cos^2 x} \text{ dx }\]


The value of \[\int\frac{\cos \sqrt{x}}{\sqrt{x}} dx\] is


The value of \[\int\frac{1}{x + x \log x} dx\] is


Evaluate the following integrals : `int (sin2x)/(cosx)dx`


Evaluate the following integrals : `int (cos2x)/(sin^2x.cos^2x)dx`


Evaluate the following integrals:

`int(2)/(sqrt(x) - sqrt(x + 3)).dx`


Integrate the following functions w.r.t.x:

`(5 - 3x)(2 - 3x)^(-1/2)`


Integrate the following functions w.r.t. x :  tan 3x tan 2x tan x


Evaluate the following : `int (1)/(4x^2 - 3).dx`


Evaluate the following : `int (1)/(5 - 4x - 3x^2).dx`


Integrate the following functions w.r.t. x : `int (1)/(3 + 2sinx).dx`


Integrate the following functions w.r.t. x : `int (1)/(3 - 2cos 2x).dx`


Integrate the following functions w.r.t. x : `int (1)/(3 + 2 sin2x + 4cos 2x).dx`


Evaluate the following integrals:

`int (2x + 1)/(x^2 + 4x - 5).dx`


Evaluate the following integrals : `int (2x + 3)/(2x^2 + 3x - 1).dx`


Evaluate the following integrals : `int sqrt((x - 7)/(x - 9)).dx`


Evaluate `int 1/(x (x - 1))` dx


Evaluate the following.

`int "x" sqrt(1 + "x"^2)` dx


Evaluate the following.

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


Evaluate the following.

∫ (x + 1)(x + 2)7 (x + 3)dx


Evaluate the following.

`int (20 - 12"e"^"x")/(3"e"^"x" - 4)`dx


Evaluate the following.

`int 1/("x"^2 + 4"x" - 5)` dx


`int x^x (1 + logx)  "d"x`


`int (cos2x)/(sin^2x)  "d"x`


`int (7x + 9)^13  "d"x` ______ + c


`int(log(logx) + 1/(logx)^2)dx` = ______.


The value of `sqrt(2) int (sinx  dx)/(sin(x - π/4))` is ______.


If `int [log(log x) + 1/(logx)^2]dx` = x [f(x) – g(x)] + C, then ______.


Evaluate the following.

`int 1/(x^2 + 4x - 5)  dx`


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


Evaluate:

`int sqrt((a - x)/x) dx`


Evaluate the following.

`intxsqrt(1+x^2)dx`


Evaluate the following

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


Evaluate the following.

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


Evaluate the following.

`int 1/ (x^2 + 4x - 5) dx`


Evaluate `int(5x^2-6x+3)/(2x-3)dx`


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


`int (x + 1)/(x(1 + xe^x)) dx` is equal to


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×