How do you integrate $\frac{1}{x - 3}$ $1/(x-3)$ using partial fractions?

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How do you integrate $\frac{1}{x - 3}$ $1/(x-3)$ using partial fractions?

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Answer 1 · 2016-08-01T15:02:40

$ln (| x - 3 |) + C$ $ln(abs(x-3))+C$

Explanation:

This cannot be split up any further into partial fractions. In fact, this is likely one of the results of a partial fraction decomposition.

When a linear factor like $x - 3$ $x-3$ is in the denominator of a fraction like this, without any other variables present, we are almost sure to use the natural logarithm integral:

$\int \frac{1}{u} d u = ln (| u |) + C$ $int1/udu=ln(absu)+C$

So here, we have:

$\int \frac{1}{x - 3} d x$ $int1/(x-3)dx$

We let: $u = x - 3$ $u=x-3$ , implying that $d u = d x$ $du=dx$ . Thus:

$\int \frac{1}{x - 3} d x = \int \frac{1}{u} d u = ln (| u |) + C = ln (| x - 3 |) + C$ $int1/(x-3)dx=int1/udu=ln(absu)+C=ln(abs(x-3))+C$

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How do you integrate $\frac{1}{x - 3}$ $1/(x-3)$ using partial fractions?

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