मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Evaluate the following integrals : ∫x-7x-9.dx

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

Let I = `int sqrt((x - 7)/(x - 9)).dx`

= `int sqrt((x - 7)/(x - 9) xx (x - 7)/(x - 7)).dx`

= `int sqrt((x - 7)^2/(x^2 - 16x + 63)).dx`

= `int (x - 7)/sqrt(x^2 - 16x + 63).dx`

Let x – 7 = `"A"[d/dx(x^2 - 16x + 63)] + "B"`

= A(2x – 16) + B
= 2Ax + (B – 16A)
Comparing the coefficient of x and constant term on both sides, we get
2A = 1

∴ A = `(1)/(2)` and

B – 16A = – 7

∴ `"B" - 16(1/2)` = – 7
∴ B = 1
∴ x – 7 = `(1)/(2)(2x - 16) + 1`

∴ I = `int (1/2(2x - 16) + 1)/sqrt(x^2 - 16x + 63).dx`

 = `(1)/(2) int (2x - 16)/sqrt(x^2 - 16x + 63).dx + int (1)/sqrt(x^2 - 16x + 63).dx`

= `(1)/(2)"I"_1 + "I"_2`

In I1, put x2 – 16x + 63 = t

∴ (2x – 16)dx = dt

∴ I1 = `(1)/(2) int (1)/sqrt(t)dt`

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

= `(1)/(2) t^(1/2)/((1/2)) + c_1`

= `sqrt(x^2 - 16x + 63) + c_1`

I2 = `int (1)/sqrt(x^2 - 16x + 63).dx`

= `int (1)/sqrt((x - 8)^2 - 1^2).dx`

= `log|x - 8 + sqrt((x - 8)^2 - 1^2)| + c_2`

= `log|x  - 8 + sqrt(x^2 - 16x + 63)| + c_2`

∴ I = `sqrt(x^2 - 16x + 63) + log|x - 8 + sqrt(x^2 - 16x + 63)| + c`,, where c = c1 + c2

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Indefinite Integration - Exercise 3.2 (C) [पृष्ठ १२८]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 3 Indefinite Integration
Exercise 3.2 (C) | Q 1.6 | पृष्ठ १२८

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Find `int((3sintheta-2)costheta)/(5-cos^2theta-4sin theta)d theta`.


Integrate the functions:

`xsqrt(x + 2)`


Integrate the functions:

`xsqrt(1+ 2x^2)`


Integrate the functions:

`sqrt(sin 2x) cos 2x`


Integrate the functions:

`(1+ log x)^2/x`


\[\int\sqrt{3 + 2x - x^2} \text{ dx}\]

\[\int e^x \sqrt{e^{2x} + 1} \text{ dx}\]

Write a value of\[\int e^{ax} \sin\ bx\ dx\]


Integrate the following w.r.t. x : x3 + x2 – x + 1


Evaluate the following integrals : `intsqrt(1 - cos 2x)dx`


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


Integrate the following functions w.r.t. x : `(1)/(4x + 5x^-11)`


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


Integrate the following functions w.r.t. x : `x^2/sqrt(9 - x^6)`


Integrate the following functions w.r.t. x : `cosx/sin(x - a)`


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


Evaluate the following : `int (1)/sqrt(3x^2 - 8).dx`


Evaluate the following : `int (1)/sqrt(11 - 4x^2).dx`


Integrate the following functions w.r.t. x : `int (1)/(4 - 5cosx).dx`


Integrate the following functions w.r.t. x : `int (1)/(cosx - sinx).dx`


Integrate the following functions w.r.t. x : `int (1)/(cosx - sqrt(3)sinx).dx`


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


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


Integrate the following with respect to the respective variable:

`x^7/(x + 1)`


Evaluate the following.

`int 1/(7 + 6"x" - "x"^2)` dx


Choose the correct alternative from the following.

The value of `int "dx"/sqrt"1 - x"` is


Fill in the Blank.

`int 1/"x"^3 [log "x"^"x"]^2 "dx" = "P" (log "x")^3` + c, then P = _______


Evaluate: `int (2"e"^"x" - 3)/(4"e"^"x" + 1)` dx


`int x^2/sqrt(1 - x^6)` dx = ________________


`int (2 + cot x - "cosec"^2x) "e"^x  "d"x`


If `tan^-1x = 2tan^-1((1 - x)/(1 + x))`, then the value of x is ______ 


`int[ tan (log x) + sec^2 (log x)] dx= ` ______


`int ("d"x)/(sinx cosx + 2cos^2x)` = ______.


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


`int (f^'(x))/(f(x))dx` = ______ + c.


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


If `int sinx/(sin^3x + cos^3x)dx = α log_e |1 + tan x| + β log_e |1 - tan x + tan^2x| + γ tan^-1 ((2tanx - 1)/sqrt(3)) + C`, when C is constant of integration, then the value of 18(α + β + γ2) is ______.


`int dx/(2 + cos x)` = ______.

(where C is a constant of integration)


Solve the following Evaluate.

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


If f ′(x) = 4x3 − 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x)


Evaluate the following.

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


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


Evaluate.

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


Evaluate the following.

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


Evaluate `int1/(x(x-1))dx`


If f '(x) = 4x3 - 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).


Evaluate the following.

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


Evaluate `int1/(x(x - 1))dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×