# How do you integrate #1/[s^2(3s+5)]# using partial fractions?

Split apart the fraction using typical decomposition rules:

Thus:

So:

Using typical integration rules (don't forget to substitute in the final integral):

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To integrate ( \frac{1}{{s^2(3s+5)}} ) using partial fractions, first express the fraction as a sum of simpler fractions:

( \frac{1}{{s^2(3s+5)}} = \frac{A}{s} + \frac{B}{{s^2}} + \frac{C}{{3s+5}} )

To solve for A, B, and C, clear the denominators:

( 1 = A(s)(3s+5) + Bs(3s+5) + Cs^2 )

Then, equate coefficients of like terms:

For ( s^0 ): ( A \times 0 = 1 \Rightarrow A = \frac{1}{5} )

For ( s^1 ): ( 3A + 3B = 0 \Rightarrow B = -\frac{1}{5} )

For ( s^2 ): ( 5A = C \Rightarrow C = \frac{3}{5} )

Now, integrate each term:

( \int \frac{1}{{s^2(3s+5)}} ds = \int \frac{1}{5s} ds - \int \frac{1}{5s^2} ds + \int \frac{3}{5(3s+5)} ds )

( = \frac{1}{5} \ln|s| + \frac{1}{5s} + \frac{1}{15} \ln|3s+5| + C )

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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