Calculus Examples

$\ln (1 + x)$

Step 1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (1 + x)$ and $d v = 1$ .

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

Step 2

Combine $x$ and $\frac{1}{x + 1}$ .

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

Step 3

Divide $x$ by $x + 1$ .

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Step 3.1

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x$

$0$

Step 3.2

Divide the highest order term in the dividend $x$ by the highest order term in divisor $x$ .

					$1$
	$x$	+	$1$		$x$	+	$0$

Step 3.3

Multiply the new quotient term by the divisor.

				$1$
$x$	+	$1$		$x$	+	$0$
			+	$x$	+	$1$

Step 3.4

The expression needs to be subtracted from the dividend, so change all the signs in $x + 1$

				$1$
$x$	+	$1$		$x$	+	$0$
			-	$x$	-	$1$

Step 3.5

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$1$
$x$	+	$1$		$x$	+	$0$
			-	$x$	-	$1$
					-	$1$

Step 3.6

The final answer is the quotient plus the remainder over the divisor.

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

Step 4

Split the single integral into multiple integrals.

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

Step 5

Apply the constant rule.

$\ln (1 + x) x - (x + C + \int - \frac{1}{x + 1} d x)$

Step 6

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$\ln (1 + x) x - (x + C - \int \frac{1}{x + 1} d x)$

Step 7

Let $u = x + 1$ . Then $d u = d x$ . Rewrite using $u$ and $d$ $u$ .

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Step 7.1

Let $u = x + 1$ . Find $\frac{d u}{d x}$ .

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Step 7.1.1

Differentiate $x + 1$ .

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

Step 7.1.2

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

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

Step 7.1.3

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

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

Step 7.1.4

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

$1 + 0$

Step 7.1.5

Add $1$ and $0$ .

$1$

Step 7.2

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

$\ln (1 + x) x - (x + C - \int \frac{1}{u} d u)$

Step 8

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

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

Step 9

Simplify.

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

Step 10

Replace all occurrences of $u$ with $x + 1$ .

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