Calculus Examples

$\frac{d y}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} + \frac{1}{x}$

Step 1

Rewrite the equation.

$d y = (\frac{- 1}{\sqrt{1 - x^{2}}} + \frac{1}{x}) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

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

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \frac{- 1}{\sqrt{1 - x^{2}}} + \frac{1}{x} d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Move the negative in front of the fraction.

$y + C_{1} = \int - \frac{1}{\sqrt{1 - x^{2}}} + \frac{1}{x} d x$

Step 2.3.2

Split the single integral into multiple integrals.

$y + C_{1} = \int - \frac{1}{\sqrt{1 - x^{2}}} d x + \int \frac{1}{x} d x$

Step 2.3.3

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

$y + C_{1} = - \int \frac{1}{\sqrt{1 - x^{2}}} d x + \int \frac{1}{x} d x$

Step 2.3.4

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

$y + C_{1} = - \int \frac{1}{\sqrt{1^{2} - x^{2}}} d x + \int \frac{1}{x} d x$

Step 2.3.5

The integral of $\frac{1}{\sqrt{1^{2} - x^{2}}}$ with respect to $x$ is $\arcsin (x)$

$y + C_{1} = - (\arcsin (x) + C_{2}) + \int \frac{1}{x} d x$

Step 2.3.6

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

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

Step 2.3.7

Simplify.

$y + C_{1} = - \arcsin (x) + \ln (| x |) + C_{4}$

Step 2.3.8

Reorder terms.

$y + C_{1} = \ln (| x |) - \arcsin (x) + C_{4}$

Step 2.4

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

$y = \ln (| x |) - \arcsin (x) + C$