English

The value of ∫23xx2+1dx is ______. - Mathematics

Advertisements
Advertisements

Question

The value of `int_2^3 x/(x^2 + 1)`dx is ______.

Options

  • `log 4`

  • `log  3/2`

  • `1/2 log2`

  • `log  9/4`

MCQ
Fill in the Blanks

Solution

The value of `int_2^3 x/(x^2 + 1)`dx is `underline(bb(1/2 log 2))`.

Explanation:

`int_2^3 x/(x^2 + 1) = 1/2 [log(x^2 + 1)]_2^3`

= `1/2 (log 10 -  log 5)`

= `1/2 log (10/5)`

= `1/2 log 2`

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
2022-2023 (March) Sample

RELATED QUESTIONS

\[\int\limits_0^{\pi/2} \cos^3 x\ dx\]

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

\[\int\limits_1^3 \frac{\cos \left( \log x \right)}{x} dx\]

\[\int\limits_0^1 \frac{\sqrt{\tan^{- 1} x}}{1 + x^2} dx\]

\[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\]

\[\int\limits_4^9 \frac{\sqrt{x}}{\left( 30 - x^{3/2} \right)^2} dx\]

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

\[\int\limits_0^\pi x \cos^2 x\ dx\]

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

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

 


\[\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .\]

 


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

Evaluate each of the following  integral:

\[\int_0^1 x e^{x^2} dx\]

 


If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.


The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is 


`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`


If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 


\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx\]  is equal to

The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .


\[\int\limits_0^4 x\sqrt{4 - x} dx\]


\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]


\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]


\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]


\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]


Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`


Using second fundamental theorem, evaluate the following:

`int_1^2 (x - 1)/x^2  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


Verify the following:

`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×