Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int\frac{1}{\sqrt{x}}\left( 1 + \frac{1}{x} \right)dx\]
\[ = \int x^{- \frac{1}{2}} \left( 1 + \frac{1}{x} \right)dx\]
\[ = \int\left( x^{- \frac{1}{2}} + \frac{1}{x^\frac{3}{2}} \right)dx\]
\[ = \int x^{- \frac{1}{2}} dx + \int x^{- \frac{3}{2}} dx\]
\[ = \left[ \frac{x^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + \left[ \frac{x^{- \frac{3}{2} + 1}}{- \frac{3}{2} + 1} \right]\]
\[ = 2\sqrt{x} - \frac{2}{\sqrt{x}} + C\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]
The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
\[\int\frac{x^3}{\sqrt{1 + x^2}}dx = a \left( 1 + x^2 \right)^\frac{3}{2} + b\sqrt{1 + x^2} + C\], 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
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
