मराठी

∫ 1 4 X 2 + 4 X + 5 D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\text{ Let I }= \int\frac{dx}{4 x^2 + 4x + 1 + 4}\]
\[ = \int\frac{dx}{\left( 2x \right)^2 + 2 \times 2x + 1 + 22}\]
\[ = \int\frac{dx}{\left( 2x + 1 \right)^2 + 2^2}\]
\[\text{ Putting }\left( 2x + 1 \right) = t\]
\[ \Rightarrow 2 \text{ dx = dt }\]
\[ \Rightarrow dx = \frac{dt}{2}\]
\[ \therefore I = \frac{1}{2}\int\frac{dt}{t^2 + 2^2}\]
\[ = \frac{1}{2} \times \frac{1}{2} \text{ tan}^{- 1} \left( \frac{t}{2} \right) + C\]
\[ = \frac{1}{4} \text{ tan}^{- 1} \left( \frac{2x + 1}{2} \right) + C ....................\left[ \because t = \left( 2x + 1 \right) \right]\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Revision Excercise [पृष्ठ २०३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Revision Excercise | Q 44 | पृष्ठ २०३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\int\left( \frac{m}{x} + \frac{x}{m} + m^x + x^m + mx \right) dx\]

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

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

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

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


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

\[\int     \text{sin}^2  \left( 2x + 5 \right)    \text{dx}\]

\[\int\frac{\sec x \tan x}{3 \sec x + 5} dx\]

\[\int\frac{\sin 2x}{\sin 5x \sin 3x} dx\]

\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]

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

\[\int \cot^5 \text{ x } {cosec}^4 x\text{ dx }\]

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

 


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

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

\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]

\[\int x e^x \text{ dx }\]

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

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

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

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

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

\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]

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

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

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

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

\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], then 


If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then


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


\[\int\sin x \sin 2x \text{ sin  3x  dx }\]


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

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

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


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

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

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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×