Calculus Examples

$\frac{d y}{d t} = \frac{y^{2} + 1}{t + 1}$

Step 1

Separate the variables.

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Step 1.1

Multiply both sides by $\frac{1}{y^{2} + 1}$ .

$\frac{1}{y^{2} + 1} \frac{d y}{d t} = \frac{1}{y^{2} + 1} \cdot \frac{y^{2} + 1}{t + 1}$

Step 1.2

Cancel the common factor of $y^{2} + 1$ .

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Step 1.2.1

Cancel the common factor.

$\frac{1}{y^{2} + 1} \frac{d y}{d t} = \frac{1}{y^{2} + 1} \cdot \frac{y^{2} + 1}{t + 1}$

Step 1.2.2

Rewrite the expression.

$\frac{1}{y^{2} + 1} \frac{d y}{d t} = \frac{1}{t + 1}$

Step 1.3

Rewrite the equation.

$\frac{1}{y^{2} + 1} d y = \frac{1}{t + 1} d t$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{1}{y^{2} + 1} d y = \int \frac{1}{t + 1} d t$

Step 2.2

Integrate the left side.

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Step 2.2.1

Simplify the expression.

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Step 2.2.1.1

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

$\int \frac{1}{1 + y^{2}} d y = \int \frac{1}{t + 1} d t$

Step 2.2.1.2

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

$\int \frac{1}{1^{2} + y^{2}} d y = \int \frac{1}{t + 1} d t$

Step 2.2.2

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

$\arctan (y) + C_{1} = \int \frac{1}{t + 1} d t$

Step 2.3

Integrate the right side.

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Step 2.3.1

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

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Step 2.3.1.1

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

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Step 2.3.1.1.1

Differentiate $t + 1$ .

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

Step 2.3.1.1.2

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

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

Step 2.3.1.1.3

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

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

Step 2.3.1.1.4

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

$1 + 0$

Step 2.3.1.1.5

Add $1$ and $0$ .

$1$

Step 2.3.1.2

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

$\arctan (y) + C_{1} = \int \frac{1}{u} d u$

Step 2.3.2

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

$\arctan (y) + C_{1} = \ln (| u |) + C_{2}$

Step 2.3.3

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

$\arctan (y) + C_{1} = \ln (| t + 1 |) + C_{2}$

Step 2.4

Group the constant of integration on the right side as $C$ .

$\arctan (y) = \ln (| t + 1 |) + C$

Step 3

Take the inverse arctangent of both sides of the equation to extract $y$ from inside the arctangent.

$y = \tan (\ln (| t + 1 |) + C)$