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

The integral of (w^2+16)/(w^2-16) is -8(1/2 ln| w/4+1|-1/2 ln| w/4-1|)+w+C

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

\int \frac{w ^{2} + 16}{w ^{2} - 16} d w

- 8 (\frac{1}{2} ln \frac{w}{4} + 1 - \frac{1}{2} ln \frac{w}{4} - 1) + w + C

Solution steps

\int \frac{w ^{2} + 16}{w ^{2} - 16} d w

\frac{w ^{2} + 16}{w ^{2} - 16} = \frac{32}{w ^{2} - 16} + 1

= \int \frac{32}{w ^{2} - 16} + 1 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{32}{w ^{2} - 16} d w + \int 1 d w

\int \frac{32}{w ^{2} - 16} d w = - 8 (\frac{1}{2} ln \frac{w}{4} + 1 - \frac{1}{2} ln \frac{w}{4} - 1)

\int 1 d w = w

= - 8 (\frac{1}{2} ln \frac{w}{4} + 1 - \frac{1}{2} ln \frac{w}{4} - 1) + w

Add a constant to the solution

= - 8 (\frac{1}{2} ln \frac{w}{4} + 1 - \frac{1}{2} ln \frac{w}{4} - 1) + w + C

\frac{1}{( s - 2 ) ( s - 4 )}

y^{^{'''}} - 12 y^{^{''}} + 52 y^{^{'}} - 96 y = 0

y = e^{x}, y = 7 e^{- x} + 6, x = 0

\int tan^{9} (x) sec^{4} (x) d x

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