Calculus Examples

$\int \frac{t^{3}}{3} + \frac{1}{4 t^{2}} d t$

Step 1

Split the single integral into multiple integrals.

$\int \frac{t^{3}}{3} d t + \int \frac{1}{4 t^{2}} d t$

Step 2

Since $\frac{1}{3}$ is constant with respect to $t$ , move $\frac{1}{3}$ out of the integral.

$\frac{1}{3} \int t^{3} d t + \int \frac{1}{4 t^{2}} d t$

Step 3

By the Power Rule, the integral of $t^{3}$ with respect to $t$ is $\frac{1}{4} t^{4}$ .

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

Step 4

Since $\frac{1}{4}$ is constant with respect to $t$ , move $\frac{1}{4}$ out of the integral.

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

Step 5

Apply basic rules of exponents.

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Step 5.1

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

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

Step 5.2

Multiply the exponents in ${(t^{2})}^{- 1}$ .

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Step 5.2.1

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

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

Step 5.2.2

Multiply $2$ by $- 1$ .

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

Step 6

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

$\frac{1}{3} (\frac{1}{4} t^{4} + C) + \frac{1}{4} (- t^{- 1} + C)$

Step 7

Simplify.

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Step 7.1

Simplify.

$\frac{1}{3} \cdot \frac{1}{4} t^{4} + \frac{1}{4} (- t^{- 1}) + C$

Step 7.2

Simplify.

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Step 7.2.1

Multiply $\frac{1}{3}$ by $\frac{1}{4}$ .

$\frac{1}{3 \cdot 4} t^{4} + \frac{1}{4} (- t^{- 1}) + C$

Step 7.2.2

Multiply $3$ by $4$ .

$\frac{1}{12} t^{4} + \frac{1}{4} (- t^{- 1}) + C$

Step 7.2.3

Combine $\frac{1}{4}$ and $t^{- 1}$ .

$\frac{1}{12} t^{4} - \frac{t^{- 1}}{4} + C$

Step 7.2.4

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

$\frac{1}{12} t^{4} - \frac{1}{4 t} + C$