CUET Preparation Today
The antiderivative of \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\) equals |
\(\frac{1}{3}x^{\frac{1}{3}}+2x^{\frac{1}{2}}+C\) \(\frac{2}{3}x^{\frac{2}{3}}+\frac{1}{2}x^{2}+C\) \(\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C\) \(\frac{3}{2}x^{\frac{3}{2}}+\frac{1}{2}x^{\frac{1}{2}}+C\) |
\(\frac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C\) |
\(\int \sqrt{x}dx=\frac{2}{3}x^{\frac{3}{2}},\int \frac{1}{\sqrt{x}}dx=2x^{\frac{1}{2}}\) |
