Number 1 $\displaystyle \int \dfrac{(w - 5) \, dw}{w^2 + 5}$
$\displaystyle = \int \left( \dfrac{w}{w^2 + 5} - \dfrac{5}{w^2 + 5} \right) \, dw$
$\displaystyle = \frac{1}{2}\int \dfrac{2w \, dw}{w^2 + 5} - 5\int \dfrac{dw}{w^2 + \left( \sqrt{5} \right)^2}$
$= \dfrac{1}{2}\ln (w^2 + 5) - \dfrac{5}{\sqrt{5}} \arctan \left( \dfrac{w}{\sqrt{5}} \right) + C$
$= \frac{1}{2}\ln (w^2 + 5) - \sqrt{5} \arctan \left( \dfrac{w}{\sqrt{5}} \right) + C$
sir question po. pano po naging 2wdw yung numerator nung 2nd part?
kasi yung derivative ng w^2+5 is 2w.. kaso w lang yung nasa numerator, so dapat lagyan ng 2 at tsaka imu-multiply by 1/2 sa labas ng integral sign para ma neutralize yung 2 sa loob..
u=w2+5 du=2wdw 1/2 du=wdw
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Number 1
$\displaystyle \int \dfrac{(w - 5) \, dw}{w^2 + 5}$
$\displaystyle = \int \left( \dfrac{w}{w^2 + 5} - \dfrac{5}{w^2 + 5} \right) \, dw$
$\displaystyle = \frac{1}{2}\int \dfrac{2w \, dw}{w^2 + 5} - 5\int \dfrac{dw}{w^2 + \left( \sqrt{5} \right)^2}$
$\displaystyle = \frac{1}{2}\int \dfrac{2w \, dw}{w^2 + 5} - 5\int \dfrac{dw}{w^2 + \left( \sqrt{5} \right)^2}$
$= \dfrac{1}{2}\ln (w^2 + 5) - \dfrac{5}{\sqrt{5}} \arctan \left( \dfrac{w}{\sqrt{5}} \right) + C$
$= \frac{1}{2}\ln (w^2 + 5) - \sqrt{5} \arctan \left( \dfrac{w}{\sqrt{5}} \right) + C$
sir question po. pano po naging 2wdw yung numerator nung 2nd part?
kasi yung derivative ng w^2+5 is 2w.. kaso w lang yung nasa numerator, so dapat lagyan ng 2 at tsaka imu-multiply by 1/2 sa labas ng integral sign para ma neutralize yung 2 sa loob..
u=w2+5
du=2wdw
1/2 du=wdw
u=w2+5
du=2wdw
1/2 du=wdw
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