Calculus Examples

$\int_{0}^{1} \frac{6 t}{t^{2} + 1} d t$

Step 1

Since $6$ is constant with respect to $t$ , move $6$ out of the integral.

$6 \int_{0}^{1} \frac{t}{t^{2} + 1} d t$

Step 2

Let $u = t^{2} + 1$ . Then $d u = 2 t d t$ , so $\frac{1}{2} d u = t d t$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.1

Let $u = t^{2} + 1$ . Find $\frac{d u}{d t}$ .

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Step 2.1.1

Differentiate $t^{2} + 1$ .

$\frac{d}{d t} [t^{2} + 1]$

Step 2.1.2

By the Sum Rule, the derivative of $t^{2} + 1$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$ .

$\frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]$

Step 2.1.3

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$2 t + \frac{d}{d t} [1]$

Step 2.1.4

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

$2 t + 0$

Step 2.1.5

Add $2 t$ and $0$ .

$2 t$

Step 2.2

Substitute the lower limit in for $t$ in $u = t^{2} + 1$ .

$u_{lower} = 0^{2} + 1$

Step 2.3

Simplify.

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Step 2.3.1

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

$u_{lower} = 0 + 1$

Step 2.3.2

Add $0$ and $1$ .

$u_{lower} = 1$

Step 2.4

Substitute the upper limit in for $t$ in $u = t^{2} + 1$ .

$u_{upper} = 1^{2} + 1$

Step 2.5

Simplify.

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Step 2.5.1

One to any power is one.

$u_{upper} = 1 + 1$

Step 2.5.2

Add $1$ and $1$ .

$u_{upper} = 2$

Step 2.6

The values found for $u_{lower}$ and $u_{upper}$ will be used to evaluate the definite integral.

$u_{lower} = 1$

$u_{upper} = 2$

Step 2.7

Rewrite the problem using $u$ , $d u$ , and the new limits of integration.

$6 \int_{1}^{2} \frac{1}{u} \cdot \frac{1}{2} d u$

Step 3

Simplify.

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Step 3.1

Multiply $\frac{1}{u}$ by $\frac{1}{2}$ .

$6 \int_{1}^{2} \frac{1}{u \cdot 2} d u$

Step 3.2

Move $2$ to the left of $u$ .

$6 \int_{1}^{2} \frac{1}{2 u} d u$

Step 4

Since $\frac{1}{2}$ is constant with respect to $u$ , move $\frac{1}{2}$ out of the integral.

$6 (\frac{1}{2} \int_{1}^{2} \frac{1}{u} d u)$

Step 5

Simplify.

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Step 5.1

Combine $\frac{1}{2}$ and $6$ .

$\frac{6}{2} \int_{1}^{2} \frac{1}{u} d u$

Step 5.2

Cancel the common factor of $6$ and $2$ .

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Step 5.2.1

Factor $2$ out of $6$ .

$\frac{2 \cdot 3}{2} \int_{1}^{2} \frac{1}{u} d u$

Step 5.2.2

Cancel the common factors.

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Step 5.2.2.1

Factor $2$ out of $2$ .

$\frac{2 \cdot 3}{2 (1)} \int_{1}^{2} \frac{1}{u} d u$

Step 5.2.2.2

Cancel the common factor.

$\frac{2 \cdot 3}{2 \cdot 1} \int_{1}^{2} \frac{1}{u} d u$

Step 5.2.2.3

Rewrite the expression.

$\frac{3}{1} \int_{1}^{2} \frac{1}{u} d u$

Step 5.2.2.4

Divide $3$ by $1$ .

$3 \int_{1}^{2} \frac{1}{u} d u$

Step 6

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

$3 (\ln (| u |)]_{1}^{2})$

Step 7

Evaluate $\ln (| u |)$ at $2$ and at $1$ .

$3 (\ln (| 2 |) - \ln (| 1 |))$

Step 8

Use the quotient property of logarithms, $\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$3 \ln (\frac{| 2 |}{| 1 |})$

Step 9

Simplify.

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Step 9.1

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$3 \ln (\frac{2}{| 1 |})$

Step 9.2

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$3 \ln (\frac{2}{1})$

Step 9.3

Divide $2$ by $1$ .

$3 \ln (2)$

Step 10

The result can be shown in multiple forms.

Exact Form:

$3 \ln (2)$

Decimal Form:

$2.07944154 \dots$

Step 11