हिंदी

1 ∫ 0 2 X 1 + X 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\text{We have}, \]
\[I = \int_0^1 \frac{2x}{1 + x^2} d x\]
\[\text{Putting} 1 + x^2 = t\]
\[ \Rightarrow 2x\ dx = dt\]
\[\text{When } x \to 0; t \to 1\]
\[\text{And } x \to 1; t \to 2\]
\[ \therefore I = \int_1^2 \frac{d t}{t}\]
\[ = \left[ \log_e \left| t \right| \right]_1^2 \]
\[ = \log_e 2 - \log_e 1\]
\[ = \log_e 2 - 0\]
\[ = \log_e 2\]

shaalaa.com
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Very Short Answers [पृष्ठ ११५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Very Short Answers | Q 25 | पृष्ठ ११५

संबंधित प्रश्न

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

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

\[\int\limits_0^{\pi/4} x^2 \sin\ x\ dx\]

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

\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]

\[\int\limits_1^2 \frac{x + 3}{x \left( x + 2 \right)} dx\]

\[\int\limits_0^1 x \left( 1 - x \right)^5 dx\]

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

\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]

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

\[\int\limits_0^1 \frac{2x}{1 + x^4} dx\]

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

 


\[\int\limits_0^{\pi/4} \sin^3 2t \cos 2t\ dt\]

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

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

Evaluate the following integral:

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

\[\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} dx\]

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

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

\[\int\limits_{- \pi/4}^{\pi/4} \sin^2 x\ dx\]

\[\int\limits_0^5 \left( x + 1 \right) dx\]

\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

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

\[\int\limits_0^{15} \left[ x \right] dx .\]

\[\int\limits_0^2 x\left[ x \right] dx .\]

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


\[\int\limits_0^1 \frac{x}{\left( 1 - x \right)^\frac{5}{4}} dx =\]

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

The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is


`int_0^(2a)f(x)dx`


\[\int\limits_1^2 x\sqrt{3x - 2} dx\]


\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]


\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]


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


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×