हिंदी

∫ Log ( Log X ) X Dx - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\text{ Let I} = \int\frac{\log \left( \log x \right) dx}{x}\]
\[\text{ Putting  log x = t}\]
\[ \Rightarrow \frac{1}{x} dx = dt\]
\[ \therefore I = \int 1_{II} \cdot \log _I  t \cdot    \text{ dt}\]
\[ = \log t\int1\text{ dt }- \int\left\{ \frac{d}{dt}\left( \log t \right)\int1 dt \right\}dt\]
\[ = \log t \cdot t - \int\frac{1}{t} \times t\text{  dt}\]
\[ = \log t \cdot t - \int dt\]
\[ = \log t \cdot t - t + C\]
\[ = t \left( \log t - 1 \right) + C\]
\[ = \log x \left( \text{ log} \left( \log x \right) - 1 \right) + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Revision Excercise [पृष्ठ २०४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Revision Excercise | Q 93 | पृष्ठ २०४

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]

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

\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]

If f' (x) = x − \[\frac{1}{x^2}\]  and  f (1)  \[\frac{1}{2},    find  f(x)\]

 


` ∫    cos  mx  cos  nx  dx `

 


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

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

\[\int 5^{x + \tan^{- 1} x} . \left( \frac{x^2 + 2}{x^2 + 1} \right) dx\]

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

\[\int \cot^n {cosec}^2 \text{ x dx } , n \neq - 1\]

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

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

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

\[\int\frac{1}{x \left( x^6 + 1 \right)} dx\]

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

`int 1/(sin x - sqrt3 cos x) dx`

\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]


`int"x"^"n"."log"  "x"  "dx"`

\[\int\left( e^\text{log  x} + \sin x \right) \text{ cos x dx }\]


\[\int x^3 \tan^{- 1}\text{  x dx }\]

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

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

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

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

\[\int\frac{x^2 + 9}{x^4 + 81} \text{ dx }\]

 


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

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

\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to

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

If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]


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

\[\int\frac{1}{\text{ sin} \left( x - a \right) \text{ sin } \left( x - b \right)} \text{ dx }\]

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


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

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

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

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

Find :  \[\int\frac{e^x}{\left( 2 + e^x \right)\left( 4 + e^{2x} \right)}dx.\] 

 


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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×