हिंदी

∫ X + 1 √ 2 X + 3 D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\int\left( \frac{x + 1}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\int\left( \frac{2x + 2}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\int\left( \frac{2x + 3 - 1}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\int\left( \frac{2x + 3}{\sqrt{2x + 3}} - \frac{1}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\int\left( \sqrt{2x + 3} - \frac{1}{\sqrt{2x + 3}} \right)dx\]
\[ = \frac{1}{2}\left[ \int \left( 2x + 3 \right)^\frac{1}{2} dx - \int \left( 2x + 3 \right)^{- \frac{1}{2}} dx \right]\]
\[ = \frac{1}{2}\left[ \frac{\left( 2x + 3 \right)^\frac{1}{2} + 1}{2\left( \frac{1}{2} + 1 \right)} - \frac{\left( 2x + 3 \right)^{- \frac{1}{2} + 1}}{2\left( - \frac{1}{2} + 1 \right)} + C \right]\]
\[ = \frac{1}{2}\left[ \frac{1}{3} \left( 2x + 3 \right)^\frac{3}{2} - \left( 2x + 3 \right)^\frac{1}{2} + C \right]\]
\[ = \frac{1}{6} \left( 2x + 3 \right)^\frac{3}{2} - \frac{1}{2} \left( 2x + 3 \right)^\frac{1}{2} + C\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.05 [पृष्ठ ३३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.05 | Q 1 | पृष्ठ ३३

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

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

` ∫ {"cosec"   x }/ { log  tan   x/2 ` dx 

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

\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]

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

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

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

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

Evaluate the following integrals:

\[\int\cos\left\{ 2 \cot^{- 1} \sqrt{\frac{1 + x}{1 - x}} \right\}dx\]

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

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

\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]

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

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

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

\[\int\frac{x^2}{x^2 + 6x + 12} \text{ dx }\]

\[\int\frac{1}{4 + 3 \tan x} dx\]

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

` ∫    sin x log  (\text{ cos x ) } dx  `

 
` ∫  x tan ^2 x dx 

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

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

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

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

\[\int\left( 2x - 5 \right) \sqrt{2 + 3x - x^2} \text{  dx }\]

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

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

\[\int\sqrt{\cot \text{θ} d  } \text{ θ}\]

\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]

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

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


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

\[\int \tan^5 x\ dx\]

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

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

\[\int \sin^{- 1} \left( 3x - 4 x^3 \right) \text{ dx}\]

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

Find: `int (3x +5)/(x^2+3x-18)dx.`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×