Advertisements
Advertisements
Question
Advertisements
Solution
\[\int\frac{dx}{\left( \sqrt{x + a} + \sqrt{x + b} \right)}\]
\[ = \int\frac{\left( \sqrt{x + a} - \sqrt{x - b} \right)}{\left( \sqrt{x + a} + \sqrt{x + b} \right)\left( \sqrt{x + a} - \sqrt{x + b} \right)}dx\]
\[ = \int\frac{\left( \sqrt{x + a} - \sqrt{x + b} \right)}{\left( x + a \right) - \left( x + b \right)}dx\]
\[ = \frac{1}{a - b}\int \left( x + a \right)^\frac{1}{2} - \frac{1}{a - b}\int \left( x + b \right)^\frac{1}{2} dx\]
` = 1/ ((a-b)) [ [(x+a ) ^ {1/2+1}] / [1/2 + 1]] ` - `1/(a-b) [[(x+b ) ^ {1/2+1}] / [1/2 + 1]] `
\[ = \frac{2}{3\left( a - b \right)}\left[ \left( x + a \right)^\frac{3}{2} - \left( x + b \right)^\frac{3}{2} \right] + C\]
APPEARS IN
RELATED QUESTIONS
\[\int\left\{ x^2 + e^{\log x}+ \left( \frac{e}{2} \right)^x \right\} dx\]
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
\[\int \left( e^x + 1 \right)^2 e^x dx\]
