What is the integral of (8w+w^2)/((w-7)(w^2+16)) ?

The integral of (8w+w^2)/((w-7)(w^2+16)) is 21/13 ln|w-7|+1/13 (-4ln|w^2+16|+12arctan(w/4))+C

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integral of (8w+w^2)/((w-7)(w^2+16))

\int \frac{8 w + w ^{2}}{( w - 7 ) ( w ^{2} + 16 )} d w

\frac{21}{13} ln ∣ w - 7 ∣ + \frac{1}{13} (- 4 ln w^{2} + 16 + 12 arctan (\frac{w}{4})) + C

Solution steps

\int \frac{8 w + w ^{2}}{( w - 7 ) ( w ^{2} + 16 )} d w

Take the partial fraction of

\frac{8 w + w ^{2}}{( w - 7 ) ( w ^{2} + 16 )} : \frac{21}{13 ( w - 7 )} + \frac{- 8 w + 48}{13 ( w ^{2} + 16 )}

= \int \frac{21}{13 ( w - 7 )} + \frac{- 8 w + 48}{13 ( w ^{2} + 16 )} d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{21}{13 ( w - 7 )} d w + \int \frac{- 8 w + 48}{13 ( w ^{2} + 16 )} d w

\int \frac{21}{13 ( w - 7 )} d w = \frac{21}{13} ln ∣ w - 7 ∣

\int \frac{- 8 w + 48}{13 ( w ^{2} + 16 )} d w = \frac{1}{13} (- 4 ln w^{2} + 16 + 12 arctan (\frac{w}{4}))

= \frac{21}{13} ln ∣ w - 7 ∣ + \frac{1}{13} (- 4 ln w^{2} + 16 + 12 arctan (\frac{w}{4}))

Add a constant to the solution

= \frac{21}{13} ln ∣ w - 7 ∣ + \frac{1}{13} (- 4 ln w^{2} + 16 + 12 arctan (\frac{w}{4})) + C

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What is the integral of (8w+w^2)/((w-7)(w^2+16)) ?
The integral of (8w+w^2)/((w-7)(w^2+16)) is 21/13 ln|w-7|+1/13 (-4ln|w^2+16|+12arctan(w/4))+C