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Integration
- Thread starter Mel Mitch
- Start date
Hello, Mel Mitch!
How about using a little algebra . . .
\(\displaystyle \text{The first integral is: }\;\int\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)\,dx\)
\(\displaystyle \text{The second is: }\;\int\left(x^{\frac{1}{2}}-x\right)\!^2\,dx \;=\;\int\left(x - 2x^{\frac{3}{2}} + x^2\right)\,dx\)
\(\displaystyle \text{Got it?}\)
How about using a little algebra . . .
\(\displaystyle \text{Integrate:}\)
\(\displaystyle 1)\;\;\int\left(x^{\frac{1}{2}}+ \frac{1}{x^{\frac{1}{2}}}\right)\,dx\)
\(\displaystyle 2)\;\;\int\left(x^{\frac{1}{2}} - x\right)\!^2\,dx\)
\(\displaystyle \text{The first integral is: }\;\int\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)\,dx\)
\(\displaystyle \text{The second is: }\;\int\left(x^{\frac{1}{2}}-x\right)\!^2\,dx \;=\;\int\left(x - 2x^{\frac{3}{2}} + x^2\right)\,dx\)
\(\displaystyle \text{Got it?}\)