English

Evaluate the following: d∫5-2x+x2dx - Mathematics

Advertisements
Advertisements

Question

Evaluate the following:

`int sqrt(5 - 2x + x^2) "d"x`

Sum
Advertisements

Solution

Let I = `int sqrt(5 - 2x + x^2) "d"x`

= `int sqrt(x^2 - 2x + 5) "d"x`

= `int sqrt(x^2 - 2x + 1 - 1 + 5) "d"x`  ....(Making perfect square)

= `int sqrt((x - 1)^2 + 4) "d"x`

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

= `(x - 1)/2 sqrt((x - 1)^2 + (2)^2) + 4/2 log|(x - 1) + sqrt((x + 1)^2 + (2)^2)| + "C"`  .......`[because int sqrt(x^2 + "a"^2) "d"x = x/2 sqrt(x^2 + "a"^2) + "a"^2/2 {log|x + sqrt(x^2 + "a"^2)|} + "C"]`

= `(x - 1)/2 sqrt(x^2 + 1 - 2x + 4) + 2log |(x - 1) + sqrt(x - 1) + sqrt(x^2 + 1 - 2x + 4)| + "C"`

= `(x - 1)/2 sqrt(x^2 - 2x + 5) + 2log|(x - 1) + sqrt(x^2 - 2x + 5)| + "C"`

Hence, I = `(x - 1)/2 sqrt(x^2 - 2x + 5) + 2log|(x - 1) + sqrt(x^2 - 2x + 5)| + "C"`

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Integrals - Exercise [Page 164]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 7 Integrals
Exercise | Q 17 | Page 164

RELATED QUESTIONS

\[\int\frac{x}{\sqrt{x + 4}} dx\]

\[\int\sqrt{\frac{1 - \cos x}{1 + \cos x}} dx\]

\[\int\frac{1 + \tan x}{1 - \tan x} dx\]

\[\int\frac{\cos 2x + x + 1}{x^2 + \sin 2x + 2x} dx\]

\[\int\frac{sec x}{\log \left( \text{sec x }+ \text{tan x} \right)} dx\]

\[\int\frac{10 x^9 + {10}^x \log_e 10}{{10}^x + x^{10}} dx\]

` ∫  {1+tan}/{ x + log  sec  x   dx} `

\[\int\frac{1}{\sin x \cos^2 x} dx\]

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

\[\int\frac{\left\{ e^{\sin^{- 1} }x \right\}^2}{\sqrt{1 - x^2}} dx\]


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


 `   ∫     tan x    .  sec^2 x   \sqrt{1 - tan^2 x}     dx\ `

Evaluate the following integrals:

\[\int\frac{1}{\left( x^2 + 2x + 10 \right)^2}dx\]

 


`  ∫    {1} / {cos x  + "cosec x" } dx  `

\[\int\frac{x + 5}{3 x^2 + 13x - 10}\text{ dx }\]

\[\int\frac{x^3 - 3x}{x^4 + 2 x^2 - 4}dx\]

\[\int\frac{1}{\sin x + \cos x} \text{ dx }\]

Evaluate the following integrals:

\[\int e^{2x} \left( \frac{1 - \sin2x}{1 - \cos2x} \right)dx\]

Evaluate the following integrals:

\[\int e^{2x} \text{ sin }\left( 3x + 1 \right) \text{ dx }\]

Evaluate the following integral:

\[\int\frac{x^3 + x + 1}{x^2 - 1}dx\]

Evaluate the following integral:

\[\int\frac{3x - 2}{\left( x + 1 \right)^2 \left( x + 3 \right)}dx\]

\[\int\frac{2x + 1}{\left( x + 2 \right) \left( x - 3 \right)^2} dx\]

Evaluate the following integral:

\[\int\frac{1}{x\left( x^3 + 8 \right)}dx\]

 


\[\int\frac{\cos x}{\left( 1 - \sin x \right) \left( 2 - \sin x \right)} dx\]

Evaluate:\[\int\frac{x^2}{1 + x^3} \text{ dx }\] .


Evaluate:  \[\int 2^x  \text{ dx }\]


Write the value of\[\int\sec x \left( \sec x + \tan x \right)\text{  dx }\]


Evaluate: \[\int\frac{x + \cos6x}{3 x^2 + \sin6x}\text{ dx }\]


Evaluate:

\[\int \cos^{-1} \left(\sin x \right) \text{dx}\]


Evaluate:

`∫ (1)/(sin^2 x cos^2 x) dx`


Evaluate : \[\int\frac{1}{x(1 + \log x)} \text{ dx}\]


Evaluate the following:

`int sqrt(2"a"x - x^2)  "d"x`


Evaluate the following:

`int sqrt(x)/(sqrt("a"^3 - x^3)) "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×