Calculus Examples

$\frac{d y}{d x} = 6 x^{- 3} + 8 x^{- 1} - 1$ ; , $y (1) = 0$

Step 1

Rewrite the equation.

$d y = (6 x^{- 3} + 8 x^{- 1} - 1) d x$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int 6 x^{- 3} + 8 x^{- 1} - 1 d x$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int 6 x^{- 3} + 8 x^{- 1} - 1 d x$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int 6 x^{- 3} d x + \int 8 x^{- 1} d x + \int - 1 d x$

Step 2.3.2

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

$y + C_{1} = 6 \int x^{- 3} d x + \int 8 x^{- 1} d x + \int - 1 d x$

Step 2.3.3

By the Power Rule, the integral of $x^{- 3}$ with respect to $x$ is $- \frac{1}{2} x^{- 2}$ .

$y + C_{1} = 6 (- \frac{1}{2} x^{- 2} + C_{2}) + \int 8 x^{- 1} d x + \int - 1 d x$

Step 2.3.4

Simplify.

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Step 2.3.4.1

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

$y + C_{1} = 6 (- \frac{x^{- 2}}{2} + C_{2}) + \int 8 x^{- 1} d x + \int - 1 d x$

Step 2.3.4.2

Move $x^{- 2}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$y + C_{1} = 6 (- \frac{1}{2 x^{2}} + C_{2}) + \int 8 x^{- 1} d x + \int - 1 d x$

Step 2.3.5

Since $8$ is constant with respect to $x$ , move $8$ out of the integral.

$y + C_{1} = 6 (- \frac{1}{2 x^{2}} + C_{2}) + 8 \int x^{- 1} d x + \int - 1 d x$

Step 2.3.6

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

$y + C_{1} = 6 (- \frac{1}{2 x^{2}} + C_{2}) + 8 (\ln (| x |) + C_{3}) + \int - 1 d x$

Step 2.3.7

Apply the constant rule.

$y + C_{1} = 6 (- \frac{1}{2 x^{2}} + C_{2}) + 8 (\ln (| x |) + C_{3}) - x + C_{4}$

Step 2.3.8

Simplify.

$y + C_{1} = - \frac{3}{x^{2}} + 8 \ln (| x |) - x + C_{5}$

Step 2.3.9

Reorder terms.

$y + C_{1} = - x - \frac{3}{x^{2}} + 8 \ln (| x |) + C_{5}$

Step 2.4

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

$y = - x - \frac{3}{x^{2}} + 8 \ln (| x |) + C$

Step 3

Use the initial condition to find the value of $C$ by substituting $1$ for $x$ and $0$ for $y$ in $y = - x - \frac{3}{x^{2}} + 8 \ln (| x |) + C$ .

$0 = - 1 - \frac{3}{1^{2}} + 8 \ln (| 1 |) + C$

Step 4

Solve for $C$ .

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Step 4.1

Rewrite the equation as $- 1 - \frac{3}{1^{2}} + 8 \ln (| 1 |) + C = 0$ .

$- 1 - \frac{3}{1^{2}} + 8 \ln (| 1 |) + C = 0$

Step 4.2

Simplify $- 1 - \frac{3}{1^{2}} + 8 \ln (| 1 |) + C$ .

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

One to any power is one.

$- 1 - \frac{3}{1} + 8 \ln (| 1 |) + C = 0$

Step 4.2.1.2

Divide $3$ by $1$ .

$- 1 - 1 \cdot 3 + 8 \ln (| 1 |) + C = 0$

Step 4.2.1.3

Multiply $- 1$ by $3$ .

$- 1 - 3 + 8 \ln (| 1 |) + C = 0$

Step 4.2.1.4

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

$- 1 - 3 + 8 \ln (1) + C = 0$

Step 4.2.1.5

The natural logarithm of $1$ is $0$ .

$- 1 - 3 + 8 \cdot 0 + C = 0$

Step 4.2.1.6

Multiply $8$ by $0$ .

$- 1 - 3 + 0 + C = 0$

Step 4.2.2

Simplify by subtracting numbers.

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Step 4.2.2.1

Add $- 1 - 3$ and $0$ .

$- 1 - 3 + C = 0$

Step 4.2.2.2

Subtract $3$ from $- 1$ .

$- 4 + C = 0$

Step 4.3

Add $4$ to both sides of the equation.

$C = 4$

Step 5

Substitute $4$ for $C$ in $y = - x - \frac{3}{x^{2}} + 8 \ln (| x |) + C$ and simplify.

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Step 5.1

Substitute $4$ for $C$ .

$y = - x - \frac{3}{x^{2}} + 8 \ln (| x |) + 4$

Step 5.2

Simplify each term.

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Step 5.2.1

Simplify $8 \ln (| x |)$ by moving $8$ inside the logarithm.

$y = - x - \frac{3}{x^{2}} + \ln ({| x |}^{8}) + 4$

Step 5.2.2

Remove the absolute value in ${| x |}^{8}$ because exponentiations with even powers are always positive.

$y = - x - \frac{3}{x^{2}} + \ln (x^{8}) + 4$