हिंदी

∫ E √ X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int e^\sqrt{x} \text{ dx }\]
योग
Advertisements

उत्तर

\[\text{ Let I } = \int e^\sqrt{x}  \text{ dx }\]
\[ = \int\sqrt{x} \cdot \frac{e^\sqrt{x}}{\sqrt{x}}dx\]
\[\text{ Let }\sqrt{x} = t\]
\[ \Rightarrow \frac{1}{2\sqrt{x}}dx = dt\]
\[ \Rightarrow \frac{dx}{\sqrt{x}} = 2 dt\]
\[ \therefore I = 2\int t_{} \cdot {e^t}_{} dt\]
`  " Taking t as the first function and e"^t" as the second function " .  `
\[ = 2\left[ t\int e^t dt - \int\left\{ \frac{d}{dt}\left( t \right)\int e^t dt \right\}dt \right] \]
\[ = 2\left[ t \cdot e^t - \int1 \cdot e^t dt \right] + C . . . (1)\]
\[\text{Substituting the value of t in eq}  \text{ (1) }\]
\[ = 2\left[ \sqrt{x} \text{ e }^\sqrt{x} - e^\sqrt{x} \right] + C\]
\[ = 2 \text{ e}^\sqrt{x} \left( \sqrt{x} - 1 \right) + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.25 [पृष्ठ १३३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.25 | Q 22 | पृष्ठ १३३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

\[\int \left( \tan x + \cot x \right)^2 dx\]

` ∫  {sec  x   "cosec " x}/{log  ( tan x) }`  dx


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

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

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

` ∫  tan 2x tan 3x  tan 5x    dx  `

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

\[\int \cot^6 x \text{ dx }\]

\[\int\frac{1}{x^2 + 6x + 13} dx\]

\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]

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

\[\int\frac{x^2}{x^2 + 7x + 10} dx\]

\[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\]

\[\int\frac{1}{13 + 3 \cos x + 4 \sin x} dx\]

\[\int\frac{3 + 2 \cos x + 4 \sin x}{2 \sin x + \cos x + 3} \text{ dx }\]

\[\int\frac{4 \sin x + 5 \cos x}{5 \sin x + 4 \cos x} \text{ dx }\]

\[\int e^x \left( \cot x - {cosec}^2 x \right) dx\]

\[\int\frac{e^x \left( x - 4 \right)}{\left( x - 2 \right)^3} \text{ dx }\]

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

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

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

Evaluate the following integral:

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

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

\[\int\frac{1}{x^4 + 3 x^2 + 1} \text{ dx }\]

\[\int\frac{1}{1 - \cos x - \sin x} dx =\]

\[\int\frac{\cos2x - \cos2\theta}{\cos x - \cos\theta}dx\] is equal to 

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

\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x}\]

\[\int \sin^3 x \cos^4 x\ \text{ dx }\]

\[\int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx }\]

\[\int\frac{1}{4 x^2 + 4x + 5} dx\]

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

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

\[\int\frac{1}{\sin^4 x + \cos^4 x} \text{ dx}\]


\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×