Evaluate integral of 1/(x-5) with respect to x

$\int \frac{1}{x - 5} d x$

Let $u = x - 5$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Let $u = x - 5$ . Find $\frac{d u}{d x}$ .

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Differentiate $x - 5$ .

$\frac{d}{d x} [x - 5]$

By the Sum Rule, the derivative of $x - 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 5]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$1 + \frac{d}{d x} [- 5]$

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5$ with respect to $x$ is $0$ .

$1 + 0$

Add $1$ and $0$ .

$1$

Rewrite the problem using $u$ and $d u$ .

$\int \frac{1}{u} d u$

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$\ln (| u |) + C$

Replace all occurrences of $u$ with $x - 5$ .

$\ln (| x - 5 |) + C$

Evaluate integral of 1/(x-5) with respect to x

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