Calculus Examples

$\int (\frac{4 + 6 h}{\sqrt{h}}) d h$

Step 1

Remove parentheses.

$\int \frac{4 + 6 h}{\sqrt{h}} d h$

Step 2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{h}$ as $h^{\frac{1}{2}}$ .

$\int \frac{4 + 6 h}{h^{\frac{1}{2}}} d h$

Step 3

Move $h^{\frac{1}{2}}$ out of the denominator by raising it to the $- 1$ power.

$\int (4 + 6 h) {(h^{\frac{1}{2}})}^{- 1} d h$

Step 4

Multiply the exponents in ${(h^{\frac{1}{2}})}^{- 1}$ .

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Step 4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int (4 + 6 h) h^{\frac{1}{2} \cdot - 1} d h$

Step 4.2

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

$\int (4 + 6 h) h^{\frac{- 1}{2}} d h$

Step 4.3

Move the negative in front of the fraction.

$\int (4 + 6 h) h^{- \frac{1}{2}} d h$

Step 5

Expand $(4 + 6 h) h^{- \frac{1}{2}}$ .

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Step 5.1

Apply the distributive property.

$\int 4 h^{- \frac{1}{2}} + 6 h \cdot h^{- \frac{1}{2}} d h$

Step 5.2

Raise $h$ to the power of $1$ .

$\int 4 h^{- \frac{1}{2}} + 6 (h^{1} h^{- \frac{1}{2}}) d h$

Step 5.3

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\int 4 h^{- \frac{1}{2}} + 6 h^{1 - \frac{1}{2}} d h$

Step 5.4

Write $1$ as a fraction with a common denominator.

$\int 4 h^{- \frac{1}{2}} + 6 h^{\frac{2}{2} - \frac{1}{2}} d h$

Step 5.5

Combine the numerators over the common denominator.

$\int 4 h^{- \frac{1}{2}} + 6 h^{\frac{2 - 1}{2}} d h$

Step 5.6

Subtract $1$ from $2$ .

$\int 4 h^{- \frac{1}{2}} + 6 h^{\frac{1}{2}} d h$

Step 5.7

Reorder $4 h^{- \frac{1}{2}}$ and $6 h^{\frac{1}{2}}$ .

$\int 6 h^{\frac{1}{2}} + 4 h^{- \frac{1}{2}} d h$

Step 6

Split the single integral into multiple integrals.

$\int 6 h^{\frac{1}{2}} d h + \int 4 h^{- \frac{1}{2}} d h$

Step 7

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

$6 \int h^{\frac{1}{2}} d h + \int 4 h^{- \frac{1}{2}} d h$

Step 8

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

$6 (\frac{2}{3} h^{\frac{3}{2}} + C) + \int 4 h^{- \frac{1}{2}} d h$

Step 9

Since $4$ is constant with respect to $h$ , move $4$ out of the integral.

$6 (\frac{2}{3} h^{\frac{3}{2}} + C) + 4 \int h^{- \frac{1}{2}} d h$

Step 10

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

$6 (\frac{2}{3} h^{\frac{3}{2}} + C) + 4 (2 h^{\frac{1}{2}} + C)$

Step 11

Simplify.

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Step 11.1

Simplify.

$6 (\frac{2}{3}) h^{\frac{3}{2}} + 4 \cdot 2 h^{\frac{1}{2}} + C$

Step 11.2

Simplify.

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Step 11.2.1

Combine $6$ and $\frac{2}{3}$ .

$\frac{6 \cdot 2}{3} h^{\frac{3}{2}} + 4 \cdot 2 h^{\frac{1}{2}} + C$

Step 11.2.2

Multiply $6$ by $2$ .

$\frac{12}{3} h^{\frac{3}{2}} + 4 \cdot 2 h^{\frac{1}{2}} + C$

Step 11.2.3

Cancel the common factor of $12$ and $3$ .

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Step 11.2.3.1

Factor $3$ out of $12$ .

$\frac{3 \cdot 4}{3} h^{\frac{3}{2}} + 4 \cdot 2 h^{\frac{1}{2}} + C$

Step 11.2.3.2

Cancel the common factors.

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Step 11.2.3.2.1

Factor $3$ out of $3$ .

$\frac{3 \cdot 4}{3 (1)} h^{\frac{3}{2}} + 4 \cdot 2 h^{\frac{1}{2}} + C$

Step 11.2.3.2.2

Cancel the common factor.

$\frac{3 \cdot 4}{3 \cdot 1} h^{\frac{3}{2}} + 4 \cdot 2 h^{\frac{1}{2}} + C$

Step 11.2.3.2.3

Rewrite the expression.

$\frac{4}{1} h^{\frac{3}{2}} + 4 \cdot 2 h^{\frac{1}{2}} + C$

Step 11.2.3.2.4

Divide $4$ by $1$ .

$4 h^{\frac{3}{2}} + 4 \cdot 2 h^{\frac{1}{2}} + C$

Step 11.2.4

Multiply $4$ by $2$ .

$4 h^{\frac{3}{2}} + 8 h^{\frac{1}{2}} + C$