Calculus Examples

$\int \frac{2 x - 1}{x (x - 1)} d x$

Step 1

Let $u = x (x - 1)$ . Then $d u = (2 x - 1) d x$ , so $\frac{1}{2 x - 1} d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 1.1

Let $u = x (x - 1)$ . Find $\frac{d u}{d x}$ .

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Step 1.1.1

Differentiate $x (x - 1)$ .

$\frac{d}{d x} [x (x - 1)]$

Step 1.1.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = x - 1$ .

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

Step 1.1.3

Differentiate.

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Step 1.1.3.1

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

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

Step 1.1.3.2

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

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

Step 1.1.3.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$x (1 + 0) + (x - 1) \frac{d}{d x} [x]$

Step 1.1.3.4

Simplify the expression.

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Step 1.1.3.4.1

Add $1$ and $0$ .

$x \cdot 1 + (x - 1) \frac{d}{d x} [x]$

Step 1.1.3.4.2

Multiply $x$ by $1$ .

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

Step 1.1.3.5

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

$x + (x - 1) \cdot 1$

Step 1.1.3.6

Simplify by adding terms.

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Step 1.1.3.6.1

Multiply $x - 1$ by $1$ .

$x + x - 1$

Step 1.1.3.6.2

Add $x$ and $x$ .

$2 x - 1$

Step 1.2

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

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

Step 2

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

$\ln (| u |) + C$

Step 3

Replace all occurrences of $u$ with $x (x - 1)$ .

$\ln (| x (x - 1) |) + C$