patulong po i solve tong mga problems (Integral Calculus)

Submitted by john.e on

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Tags
Integration
Logarithmic Function
Inverse Trigonometric Function

Jhun Vert Thu, 12/10/2015 - 06:29

Number 1
$\displaystyle \int \dfrac{(w - 5) \, dw}{w^2 + 5}$

      $\displaystyle = \int \left( \dfrac{w}{w^2 + 5} - \dfrac{5}{w^2 + 5} \right) \, dw$

      $\displaystyle = \frac{1}{2}\int \dfrac{2w \, dw}{w^2 + 5} - 5\int \dfrac{dw}{w^2 + \left( \sqrt{5} \right)^2}$

      $\displaystyle = \frac{1}{2}\int \dfrac{2w \, dw}{w^2 + 5} - 5\int \dfrac{dw}{w^2 + \left( \sqrt{5} \right)^2}$

      $= \dfrac{1}{2}\ln (w^2 + 5) - \dfrac{5}{\sqrt{5}} \arctan \left( \dfrac{w}{\sqrt{5}} \right) + C$

      $= \frac{1}{2}\ln (w^2 + 5) - \sqrt{5} \arctan \left( \dfrac{w}{\sqrt{5}} \right) + C$