मराठी

∫ X + 1 X ( X + Log X ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

` Note: "Here, we are considering " log x as log_e x `
\[\text{Let I} = \int\frac{x + 1}{x\left( x + \ logx \right)}dx\]
\[\text{Putting}\ x + \log x = t\]
\[ \Rightarrow 1 + \frac{1}{x} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{x + 1}{x}dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{log }\left| t \right| + C\]
\[ = \text{log }\left| x + \log x \right| + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.08 [पृष्ठ ४८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.08 | Q 39 | पृष्ठ ४८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\int\left( x^e + e^x + e^e \right) dx\]

\[\int\left( \sec^2  x + {cosec}^2  x \right)  dx\]

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

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

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

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

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

` ∫   tan   x   sec^4  x   dx  `


` ∫      tan^5    x   dx `


\[\int \cot^5 x  \text{ dx }\]

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

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

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

\[\int\frac{1}{\sqrt{16 - 6x - x^2}} dx\]

` ∫  {x-3} /{ x^2 + 2x - 4 } dx `


\[\int\frac{x + 7}{3 x^2 + 25x + 28}\text{ dx}\]

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

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

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

\[\int\frac{1}{p + q \tan x} \text{ dx  }\]

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

\[\int x^3 \tan^{- 1}\text{  x dx }\]

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

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

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

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

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

\[\int\frac{x}{4 + x^4} \text{ dx }\] is equal to

\[\int\sqrt{\frac{x}{1 - x}} dx\]  is equal to


\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\]  is equal to 

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

\[\int x \sin^5 x^2 \cos x^2 dx\]

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


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

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

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

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

\[\int\frac{\sin 4x - 2}{1 - \cos 4x} e^{2x} \text{ dx}\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×