हिंदी

∫ 1 √ X + a + √ X + B D X

Advertisements
Advertisements

प्रश्न

\[\int\frac{1}{\sqrt{x + a} + \sqrt{x + b}} dx\]
योग
Advertisements

उत्तर

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

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 18: Indefinite Integrals - Exercise 19.03 [पृष्ठ २३]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 18 Indefinite Integrals
Exercise 19.03 | Q 8 | पृष्ठ २३
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×