हिंदी

∫ E X X { X ( Log X ) 2 + 2 Log X } D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{e^x}{x}\left\{ \text{ x }\left( \log x \right)^2 + 2 \log x \right\} dx\]
योग
Advertisements

उत्तर

\[\text{ Let I }= \int\frac{e^x}{x}\left[ x \left( \log x \right)^2 + 2\log x \right]dx\]

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

\[Here, f(x) = \left( \log x \right)^2 \]

\[ \Rightarrow f'(x) = \frac{2\log x}{x}\]

\[\text{ put  e}^x f(x) = t\]

\[ \Rightarrow e^x \left( \log x \right)^2 = t\]

\[\text{ Diff   both    sides   w . r . t x }\]

\[\left[ e^x \left( \log  x \right)^2 + e^x \frac{2\log x}{x} \right]dx = dt\]

\[ \therefore I = \int dt\]

\[ = t + C\]

\[ = e^x \left( \log x \right)^2 + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.26 [पृष्ठ १४३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.26 | Q 17 | पृष्ठ १४३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

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

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

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

\[\int\frac{1 + \cos 4x}{\cot x - \tan x} dx\]

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

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

\[\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx\]

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

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

\[\ \int\ x \left( 1 - x \right)^{23} dx\]

 


\[\int \sin^3 x \cos^5 x \text{ dx  }\]

Evaluate the following integrals:
\[\int\frac{x^2}{\left( a^2 - x^2 \right)^{3/2}}dx\]

\[\int\frac{1}{\sqrt{a^2 + b^2 x^2}} dx\]

\[\int\frac{1}{\sqrt{3 x^2 + 5x + 7}} dx\]

\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]

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

\[\int\frac{5x + 3}{\sqrt{x^2 + 4x + 10}} \text{ dx }\]

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

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

\[\int\frac{1}{1 - \cot x} dx\]

 
` ∫  x tan ^2 x dx 

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

\[\int x^3 \tan^{- 1}\text{  x dx }\]

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

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

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

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

\[\int\frac{x^2}{\left( x^2 + 1 \right) \left( 3 x^2 + 4 \right)} dx\]

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

\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to

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

\[\int x\sqrt{2x + 3} \text{ dx }\]

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

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

\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]


\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]


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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×