Calculus Examples

$\int_{0}^{1} \ln (x^{2} + 1) d x$

Step 1

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

$\ln (x^{2} + 1) x]_{0}^{1} - \int_{0}^{1} x \frac{2 x}{x^{2} + 1} d x$

Step 2

Simplify.

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Step 2.1

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

$\ln (x^{2} + 1) x]_{0}^{1} - \int_{0}^{1} \frac{x (2 x)}{x^{2} + 1} d x$

Step 2.2

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

$\ln (x^{2} + 1) x]_{0}^{1} - \int_{0}^{1} \frac{2 (x^{1} x)}{x^{2} + 1} d x$

Step 2.3

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

$\ln (x^{2} + 1) x]_{0}^{1} - \int_{0}^{1} \frac{2 (x^{1} x^{1})}{x^{2} + 1} d x$

Step 2.4

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

$\ln (x^{2} + 1) x]_{0}^{1} - \int_{0}^{1} \frac{2 x^{1 + 1}}{x^{2} + 1} d x$

Step 2.5

Add $1$ and $1$ .

$\ln (x^{2} + 1) x]_{0}^{1} - \int_{0}^{1} \frac{2 x^{2}}{x^{2} + 1} d x$

Step 3

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

$\ln (x^{2} + 1) x]_{0}^{1} - (2 \int_{0}^{1} \frac{x^{2}}{x^{2} + 1} d x)$

Step 4

Multiply $2$ by $- 1$ .

$\ln (x^{2} + 1) x]_{0}^{1} - 2 \int_{0}^{1} \frac{x^{2}}{x^{2} + 1} d x$

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.

$\ln (x^{2} + 1) x]_{0}^{1} - 2 \int_{0}^{1} 1 - \frac{1}{x^{2} + 1} d x$

Step 6

Split the single integral into multiple integrals.

$\ln (x^{2} + 1) x]_{0}^{1} - 2 (\int_{0}^{1} d x + \int_{0}^{1} - \frac{1}{x^{2} + 1} d x)$

Step 7

Apply the constant rule.

$\ln (x^{2} + 1) x]_{0}^{1} - 2 (x]_{0}^{1} + \int_{0}^{1} - \frac{1}{x^{2} + 1} d x)$

Step 8

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

$\ln (x^{2} + 1) x]_{0}^{1} - 2 (x]_{0}^{1} - \int_{0}^{1} \frac{1}{x^{2} + 1} d x)$

Step 9

Simplify the expression.

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Step 9.1

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

$\ln (x^{2} + 1) x]_{0}^{1} - 2 (x]_{0}^{1} - \int_{0}^{1} \frac{1}{1 + x^{2}} d x)$

Step 9.2

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

$\ln (x^{2} + 1) x]_{0}^{1} - 2 (x]_{0}^{1} - \int_{0}^{1} \frac{1}{1^{2} + x^{2}} d x)$

Step 10

The integral of $\frac{1}{1^{2} + x^{2}}$ with respect to $x$ is $\arctan (x)]_{0}^{1}$ .

$\ln (x^{2} + 1) x]_{0}^{1} - 2 (x]_{0}^{1} - (\arctan (x)]_{0}^{1}))$

Step 11

Substitute and simplify.

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Step 11.1

Evaluate $\ln (x^{2} + 1) x$ at $1$ and at $0$ .

$(\ln (1^{2} + 1) \cdot 1) - \ln (0^{2} + 1) \cdot 0 - 2 (x]_{0}^{1} - (\arctan (x)]_{0}^{1}))$

Step 11.2

Evaluate $x$ at $1$ and at $0$ .

$(\ln (1^{2} + 1) \cdot 1) - \ln (0^{2} + 1) \cdot 0 - 2 ((1) + 0 - (\arctan (x)]_{0}^{1}))$

Step 11.3

Evaluate $\arctan (x)$ at $1$ and at $0$ .

$(\ln (1^{2} + 1) \cdot 1) - \ln (0^{2} + 1) \cdot 0 - 2 (1 + 0 - (\arctan (1) - \arctan (0)))$

Step 11.4

Simplify.

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Step 11.4.1

One to any power is one.

$\ln (1 + 1) \cdot 1 - \ln (0^{2} + 1) \cdot 0 - 2 (1 + 0 - (\arctan (1) - \arctan (0)))$

Step 11.4.2

Add $1$ and $1$ .

$\ln (2) \cdot 1 - \ln (0^{2} + 1) \cdot 0 - 2 (1 + 0 - (\arctan (1) - \arctan (0)))$

Step 11.4.3

Multiply $\ln (2)$ by $1$ .

$\ln (2) - \ln (0^{2} + 1) \cdot 0 - 2 (1 + 0 - (\arctan (1) - \arctan (0)))$

Step 11.4.4

Raising $0$ to any positive power yields $0$ .

$\ln (2) - \ln (0 + 1) \cdot 0 - 2 (1 + 0 - (\arctan (1) - \arctan (0)))$

Step 11.4.5

Add $0$ and $1$ .

$\ln (2) - \ln (1) \cdot 0 - 2 (1 + 0 - (\arctan (1) - \arctan (0)))$

Step 11.4.6

Multiply $0$ by $- 1$ .

$\ln (2) + 0 \ln (1) - 2 (1 + 0 - (\arctan (1) - \arctan (0)))$

Step 11.4.7

Multiply $0$ by $\ln (1)$ .

$\ln (2) + 0 - 2 (1 + 0 - (\arctan (1) - \arctan (0)))$

Step 11.4.8

Add $\ln (2)$ and $0$ .

$\ln (2) - 2 (1 + 0 - (\arctan (1) - \arctan (0)))$

Step 11.4.9

Add $1$ and $0$ .

$\ln (2) - 2 (1 - (\arctan (1) - \arctan (0)))$

Step 12

Simplify each term.

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Step 12.1

Simplify each term.

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Step 12.1.1

Simplify each term.

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Step 12.1.1.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$\ln (2) - 2 (1 - (\frac{π}{4} - \arctan (0)))$

Step 12.1.1.2

The exact value of $\arctan (0)$ is $0$ .

$\ln (2) - 2 (1 - (\frac{π}{4} - 0))$

Step 12.1.1.3

Multiply $- 1$ by $0$ .

$\ln (2) - 2 (1 - (\frac{π}{4} + 0))$

Step 12.1.2

Add $\frac{π}{4}$ and $0$ .

$\ln (2) - 2 (1 - \frac{π}{4})$

Step 12.2

Apply the distributive property.

$\ln (2) - 2 \cdot 1 - 2 (- \frac{π}{4})$

Step 12.3

Multiply $- 2$ by $1$ .

$\ln (2) - 2 - 2 (- \frac{π}{4})$

Step 12.4

Cancel the common factor of $2$ .

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Step 12.4.1

Move the leading negative in $- \frac{π}{4}$ into the numerator.

$\ln (2) - 2 - 2 \frac{- π}{4}$

Step 12.4.2

Factor $2$ out of $- 2$ .

$\ln (2) - 2 + 2 (- 1) \frac{- π}{4}$

Step 12.4.3

Factor $2$ out of $4$ .

$\ln (2) - 2 + 2 \cdot - 1 \frac{- π}{2 \cdot 2}$

Step 12.4.4

Cancel the common factor.

$\ln (2) - 2 + 2 \cdot - 1 \frac{- π}{2 \cdot 2}$

Step 12.4.5

Rewrite the expression.

$\ln (2) - 2 - \frac{- π}{2}$

Step 12.5

Simplify each term.

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Step 12.5.1

Move the negative in front of the fraction.

$\ln (2) - 2 - - \frac{π}{2}$

Step 12.5.2

Multiply $- - \frac{π}{2}$ .

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Step 12.5.2.1

Multiply $- 1$ by $- 1$ .

$\ln (2) - 2 + 1 \frac{π}{2}$

Step 12.5.2.2

Multiply $\frac{π}{2}$ by $1$ .

$\ln (2) - 2 + \frac{π}{2}$

Step 13

The result can be shown in multiple forms.

Exact Form:

$\ln (2) - 2 + \frac{π}{2}$

Decimal Form:

$0.26394350 \dots$