Calculus Examples

$\frac{3 y^{2} + 1}{y^{3} + y} \cdot \frac{d y}{d x} = 2 x$

Step 1

Rewrite the equation.

$\frac{3 y^{2} + 1}{y^{3} + y} \cdot d y = 2 x d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int \frac{3 y^{2} + 1}{y^{3} + y} d y = \int 2 x d x$

Step 2.2

Integrate the left side.

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Step 2.2.1

Let $u = y^{3} + y$ . Then $d u = (3 y^{2} + 1) d y$ , so $\frac{1}{3 y^{2} + 1} d u = d y$ . Rewrite using $u$ and $d$ $u$ .

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Step 2.2.1.1

Let $u = y^{3} + y$ . Find $\frac{d u}{d y}$ .

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Step 2.2.1.1.1

Differentiate $y^{3} + y$ .

$\frac{d}{d y} [y^{3} + y]$

Step 2.2.1.1.2

By the Sum Rule, the derivative of $y^{3} + y$ with respect to $y$ is $\frac{d}{d y} [y^{3}] + \frac{d}{d y} [y]$ .

$\frac{d}{d y} [y^{3}] + \frac{d}{d y} [y]$

Step 2.2.1.1.3

Differentiate using the Power Rule which states that $\frac{d}{d y} [y^{n}]$ is $n y^{n - 1}$ where $n = 3$ .

$3 y^{2} + \frac{d}{d y} [y]$

Step 2.2.1.1.4

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

$3 y^{2} + 1$

Step 2.2.1.2

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

$\int \frac{1}{u} d u = \int 2 x d x$

Step 2.2.2

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

$\ln (| u |) + C_{1} = \int 2 x d x$

Step 2.2.3

Replace all occurrences of $u$ with $y^{3} + y$ .

$\ln (| y^{3} + y |) + C_{1} = \int 2 x d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

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

$\ln (| y^{3} + y |) + C_{1} = 2 \int x d x$

Step 2.3.2

By the Power Rule, the integral of $x$ with respect to $x$ is $\frac{1}{2} x^{2}$ .

$\ln (| y^{3} + y |) + C_{1} = 2 (\frac{1}{2} x^{2} + C_{2})$

Step 2.3.3

Simplify the answer.

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Step 2.3.3.1

Rewrite $2 (\frac{1}{2} x^{2} + C_{2})$ as $2 (\frac{1}{2}) x^{2} + C_{2}$ .

$\ln (| y^{3} + y |) + C_{1} = 2 (\frac{1}{2}) x^{2} + C_{2}$

Step 2.3.3.2

Simplify.

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Step 2.3.3.2.1

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

$\ln (| y^{3} + y |) + C_{1} = \frac{2}{2} x^{2} + C_{2}$

Step 2.3.3.2.2

Cancel the common factor of $2$ .

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Step 2.3.3.2.2.1

Cancel the common factor.

$\ln (| y^{3} + y |) + C_{1} = \frac{2}{2} x^{2} + C_{2}$

Step 2.3.3.2.2.2

Rewrite the expression.

$\ln (| y^{3} + y |) + C_{1} = 1 x^{2} + C_{2}$

Step 2.3.3.2.3

Multiply $x^{2}$ by $1$ .

$\ln (| y^{3} + y |) + C_{1} = x^{2} + C_{2}$

Step 2.4

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

$\ln (| y^{3} + y |) = x^{2} + C$