Calculus Examples

$\frac{d x}{d t} = 4 (x^{2} + 1)$

Step 1

Separate the variables.

Tap for more steps...

Step 1.1

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

$\frac{1}{x^{2} + 1} \frac{d x}{d t} = \frac{1}{x^{2} + 1} (4 (x^{2} + 1))$

Step 1.2

Simplify.

Tap for more steps...

Step 1.2.1

Rewrite using the commutative property of multiplication.

$\frac{1}{x^{2} + 1} \frac{d x}{d t} = 4 \frac{1}{x^{2} + 1} (x^{2} + 1)$

Step 1.2.2

Combine $4$ and $\frac{1}{x^{2} + 1}$ .

$\frac{1}{x^{2} + 1} \frac{d x}{d t} = \frac{4}{x^{2} + 1} (x^{2} + 1)$

Step 1.2.3

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

Tap for more steps...

Step 1.2.3.1

Cancel the common factor.

$\frac{1}{x^{2} + 1} \frac{d x}{d t} = \frac{4}{x^{2} + 1} (x^{2} + 1)$

Step 1.2.3.2

Rewrite the expression.

$\frac{1}{x^{2} + 1} \frac{d x}{d t} = 4$

Step 1.3

Rewrite the equation.

$\frac{1}{x^{2} + 1} d x = 4 d t$

Step 2

Integrate both sides.

Tap for more steps...

Step 2.1

Set up an integral on each side.

$\int \frac{1}{x^{2} + 1} d x = \int 4 d t$

Step 2.2

Integrate the left side.

Tap for more steps...

Step 2.2.1

Simplify the expression.

Tap for more steps...

Step 2.2.1.1

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

$\int \frac{1}{1 + x^{2}} d x = \int 4 d t$

Step 2.2.1.2

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

$\int \frac{1}{1^{2} + x^{2}} d x = \int 4 d t$

Step 2.2.2

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

$\arctan (x) + C_{1} = \int 4 d t$

Step 2.3

Apply the constant rule.

$\arctan (x) + C_{1} = 4 t + C_{2}$

Step 2.4

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

$\arctan (x) = 4 t + K$

Step 3

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

$x = \tan (4 t + K)$