Calculus Examples

$\int \ln (x^{2} + 16) d x$

Step 1

Integrate by parts using the formula $\int u d v = u v - \int v d u$ , where $u = \ln (x^{2} + 16)$ and $d v = 1$ .

$\ln (x^{2} + 16) x - \int x \frac{2 x}{x^{2} + 16} d x$

Step 2

Simplify.

Tap for more steps...

Step 2.1

Combine $x$ and $\frac{2 x}{x^{2} + 16}$ .

$\ln (x^{2} + 16) x - \int \frac{x (2 x)}{x^{2} + 16} d x$

Step 2.2

Raise $x$ to the power of $1$ .

$\ln (x^{2} + 16) x - \int \frac{2 (x^{1} x)}{x^{2} + 16} d x$

Step 2.3

Raise $x$ to the power of $1$ .

$\ln (x^{2} + 16) x - \int \frac{2 (x^{1} x^{1})}{x^{2} + 16} d x$

Step 2.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\ln (x^{2} + 16) x - \int \frac{2 x^{1 + 1}}{x^{2} + 16} d x$

Step 2.5

Add $1$ and $1$ .

$\ln (x^{2} + 16) x - \int \frac{2 x^{2}}{x^{2} + 16} d x$

Step 3

Since $2$ is constant with respect to $x$ , move $2$ out of the integral.

$\ln (x^{2} + 16) x - (2 \int \frac{x^{2}}{x^{2} + 16} d x)$

Step 4

Multiply $2$ by $- 1$ .

$\ln (x^{2} + 16) x - 2 \int \frac{x^{2}}{x^{2} + 16} d x$

Step 5

Divide $x^{2}$ by $x^{2} + 16$ .

Tap for more steps...

Step 5.1

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

$x^{2}$

$0 x$

$16$

$x^{2}$

$0 x$

$0$

Step 5.2

Divide the highest order term in the dividend $x^{2}$ by the highest order term in divisor $x^{2}$ .

							$1$
	$x^{2}$	+	$0 x$	+	$16$		$x^{2}$	+	$0 x$	+	$0$

Step 5.3

Multiply the new quotient term by the divisor.

						$1$
$x^{2}$	+	$0 x$	+	$16$		$x^{2}$	+	$0 x$	+	$0$
					+	$x^{2}$	+	$0$	+	$16$

Step 5.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{2} + 0 + 16$

						$1$
$x^{2}$	+	$0 x$	+	$16$		$x^{2}$	+	$0 x$	+	$0$
					-	$x^{2}$	-	$0$	-	$16$

Step 5.5

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

						$1$
$x^{2}$	+	$0 x$	+	$16$		$x^{2}$	+	$0 x$	+	$0$
					-	$x^{2}$	-	$0$	-	$16$
									-	$16$

Step 5.6

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

$\ln (x^{2} + 16) x - 2 \int 1 - \frac{16}{x^{2} + 16} d x$

Step 6

Split the single integral into multiple integrals.

$\ln (x^{2} + 16) x - 2 (\int d x + \int - \frac{16}{x^{2} + 16} d x)$

Step 7

Apply the constant rule.

$\ln (x^{2} + 16) x - 2 (x + C + \int - \frac{16}{x^{2} + 16} d x)$

Step 8

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

$\ln (x^{2} + 16) x - 2 (x + C - \int \frac{16}{x^{2} + 16} d x)$

Step 9

Since $16$ is constant with respect to $x$ , move $16$ out of the integral.

$\ln (x^{2} + 16) x - 2 (x + C - (16 \int \frac{1}{x^{2} + 16} d x))$

Step 10

Simplify the expression.

Tap for more steps...

Step 10.1

Multiply $16$ by $- 1$ .

$\ln (x^{2} + 16) x - 2 (x + C - 16 \int \frac{1}{x^{2} + 16} d x)$

Step 10.2

Reorder $x^{2}$ and $16$ .

$\ln (x^{2} + 16) x - 2 (x + C - 16 \int \frac{1}{16 + x^{2}} d x)$

Step 10.3

Rewrite $16$ as $4^{2}$ .

$\ln (x^{2} + 16) x - 2 (x + C - 16 \int \frac{1}{4^{2} + x^{2}} d x)$

Step 11

The integral of $\frac{1}{4^{2} + x^{2}}$ with respect to $x$ is $\frac{1}{4} \arctan (\frac{x}{4}) + C$ .

$\ln (x^{2} + 16) x - 2 (x + C - 16 (\frac{1}{4} \arctan (\frac{x}{4}) + C))$

Step 12

Simplify.

Tap for more steps...

Step 12.1

Combine $\frac{1}{4}$ and $\arctan (\frac{x}{4})$ .

$\ln (x^{2} + 16) x - 2 (x + C - 16 (\frac{\arctan (\frac{x}{4})}{4} + C))$

Step 12.2

Simplify.

$\ln (x^{2} + 16) x - 2 (x - 4 \arctan (\frac{x}{4})) + C$

Step 12.3

Reorder terms.

$\ln (x^{2} + 16) x - 2 (x - 4 \arctan (\frac{1}{4} x)) + C$