हिंदी

Ed∫ex(1-x1+x2)2 dx is equal to ______.

Advertisements
Advertisements

प्रश्न

`int "e"^x ((1 - x)/(1 + x^2))^2  "d"x` is equal to ______.

विकल्प

  • `"e"^x/(1 + x^2) + "C"`

  • `(-"e"^x)/(1 + x^2) + "C"`

  • `"e"^x/(1 + x^2)^2 + "C"`

  • `(-"e"^x)/(1 + x^2)^2 + "C"`

MCQ
रिक्त स्थान भरें
Advertisements

उत्तर

`int "e"^x ((1 - x)/(1 + x^2))^2  "d"x` is equal to `"e"^x/(1 + x^2) + "C"`.

Explanation:

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

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

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

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

Here f(x) = `1/(1 + x^2)`

∴ f'(x) = `(-2x)/(1 + x^2)^2`

Using `int "e"^x ["f"(x) + "f'"(x)]"d"x = "e"^x * "f"(x) + "C"`

∴ I = `"e"^x * 1/(1 + x^2) + "C" = "e"^x/(1 + x^2) + "C"`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Integrals - Exercise [पृष्ठ १६७]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 7 Integrals
Exercise | Q 51 | पृष्ठ १६७

संबंधित प्रश्न

\[\int\limits_0^\infty e^{- x} dx\]

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

\[\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx\]

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

\[\int\limits_1^e \frac{e^x}{x} \left( 1 + x \log x \right) dx\]

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

\[\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx\]

\[\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx\]

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

\[\int\limits_0^1 \frac{e^x}{1 + e^{2x}} dx\]

\[\int\limits_0^1 x e^{x^2} dx\]

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 


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

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

\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]

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

\[\int\limits_0^a \sin^{- 1} \sqrt{\frac{x}{a + x}} dx\]

\[\int_0^\frac{\pi}{2} \frac{\tan x}{1 + m^2 \tan^2 x}dx\]

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

Evaluate each of the following integral:

\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]


\[\int\limits_0^2 x\sqrt{2 - x} dx\]

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

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

 


\[\int\limits_1^4 \left( x^2 - x \right) dx\]

\[\int\limits_0^3 \frac{1}{x^2 + 9} dx .\]

If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.

 

 


Write the coefficient abc of which the value of the integral

\[\int\limits_{- 3}^3 \left( a x^2 + bx + c \right) dx\] is independent.

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


\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals


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

 


The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is 


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


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


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


\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]


Using second fundamental theorem, evaluate the following:

`int_0^1 x"e"^(x^2)  "d"x`


Evaluate the following:

`int_0^oo "e"^(-mx) x^6 "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 x^2  "d"x`


Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`


If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.


Find: `int logx/(1 + log x)^2 dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×