Calculus Examples

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

Step 1

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

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

Step 2

Simplify.

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Step 2.1

Combine $x$ and $\frac{2 x}{x^{2} + 1}$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Add $1$ and $1$ .

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

Step 3

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

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

Step 4

Multiply $2$ by $- 1$ .

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

Step 5

Divide $x^{2}$ by $x^{2} + 1$ .

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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 $0$ .

$x^{2}$

$0 x$

$1$

$x^{2}$

$0 x$

$0$

Step 5.2

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

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

Step 5.3

Multiply the new quotient term by the divisor.

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

Step 5.4

The expression needs to be subtracted from the dividend, so change all the signs in $x^{2} + 0 + 1$

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

Step 5.5

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

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

Step 5.6

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

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

Step 6

Split the single integral into multiple integrals.

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

Step 7

Apply the constant rule.

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

Step 8

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

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

Step 9

Simplify the expression.

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Step 9.1

Reorder $x^{2}$ and $1$ .

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

Step 9.2

Rewrite $1$ as $1^{2}$ .

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

Step 10

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

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

Step 11

Simplify.

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