English

D∫x9(4x2+1)6 dx is equal to ______. - Mathematics

Advertisements
Advertisements

Question

`int x^9/(4x^2 + 1)^6  "d"x` is equal to ______.

Options

  • `1/(5x)(4 + 1/x^2)^-5 + "C"`

  • `1/5(4 + 1/x^2)^-5 + "C"`

  • `1/(10x)(1 + 4)^-5 + "C"`

  • `1/10(1/x^2 + 4)^-5 + "C"`

MCQ
Fill in the Blanks
Advertisements

Solution

`int x^9/(4x^2 + 1)^6  "d"x` is equal to `1/10(1/x^2 + 4)^-5 + "C"`.

Explanation:

Let I = `int x^9/(4x^2 + 1)^6 "d"x`

= `int  x^9/(x^12(4 + 1/x^2)^6) "d"x`

= `int 1/(x^3(4 + 1/x^2)^6) "d"x`

Put `(4 + 1/x^2)` = t

⇒ `(-2)/x^3 "dt"` = dt

⇒ `"dx"/x^3 = - 1/2 "dt"`

∴ I = `- 1/2 int "dt"/"t"^6`

= `- 1/2 xx - 1/5 "t"^-5 + "C"`

= `1/10 "t"^-5 + "C"`

= `1/10(4 + 1/x^2)^-5 + "C"`

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 7: Integrals - Exercise [Page 167]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 7 Integrals
Exercise | Q 52 | Page 167

RELATED QUESTIONS

\[\int\limits_{- 2}^3 \frac{1}{x + 7} dx\]

\[\int\limits_0^1 \frac{x}{x + 1} dx\]

\[\int\limits_0^{\pi/4} \sec x dx\]

\[\int\limits_0^{\pi/2} \sin x \sin 2x\ dx\]

\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]

\[\int\limits_0^{\pi/2} x \cos\ x\ dx\]

\[\int\limits_1^2 \log\ x\ dx\]

\[\int\limits_{- 1}^1 \frac{1}{x^2 + 2x + 5} dx\]

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx\]

\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]

Evaluate the following integral:

\[\int\limits_{- 2}^2 \left| 2x + 3 \right| dx\]

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) dx\]

If f is an integrable function, show that

\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]


\[\int\limits_a^b x\ dx\]

\[\int\limits_0^{\pi/2} \sin^2 x\ dx .\]

\[\int\limits_{- \pi/2}^{\pi/2} \cos^2 x\ dx .\]

\[\int\limits_0^{\pi/2} \log \tan x\ dx .\]

\[\int\limits_0^\pi \cos^5 x\ dx .\]

If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:

\[\int\limits_0^{\pi/4} \sin \left\{ x \right\} dx\]

 


The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is

 


If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals


\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^3 x} dx\]  is equal to

\[\int\limits_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx\]  equals to

\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\]  is equal to 

\[\int\limits_0^1 \cos^{- 1} x dx\]


\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]


\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]


Evaluate the following integrals :-

\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]


\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]


\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]


\[\int\limits_0^\pi x \sin x \cos^4 x dx\]


\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]


\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]


Using second fundamental theorem, evaluate the following:

`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`


Evaluate the following using properties of definite integral:

`int_(- pi/4)^(pi/4) x^3 cos^3 x  "d"x`


Evaluate the following:

`Γ (9/2)`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×