Calculus Examples

$\int \frac{4 x}{e^{5 x^{2}}} d x$

Step 1

Let $u_{1} = 5 x^{2}$ . Then $d u_{1} = 10 x d x$ , so $\frac{1}{10} d u_{1} = x d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 1.1

Let $u_{1} = 5 x^{2}$ . Find $\frac{d u_{1}}{d x}$ .

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Step 1.1.1

Differentiate $5 x^{2}$ .

$\frac{d}{d x} [5 x^{2}]$

Step 1.1.2

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2}$ with respect to $x$ is $5 \frac{d}{d x} [x^{2}]$ .

$5 \frac{d}{d x} [x^{2}]$

Step 1.1.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$5 (2 x)$

Step 1.1.4

Multiply $2$ by $5$ .

$10 x$

Step 1.2

Rewrite the problem using $u_{1}$ and $d u_{1}$ .

$\int \frac{4}{e^{u_{1}}} \cdot \frac{1}{10} d u_{1}$

Step 2

Simplify.

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Step 2.1

Multiply $\frac{4}{e^{u_{1}}}$ by $\frac{1}{10}$ .

$\int \frac{4}{e^{u_{1}} \cdot 10} d u_{1}$

Step 2.2

Move $10$ to the left of $e^{u_{1}}$ .

$\int \frac{4}{10 \cdot e^{u_{1}}} d u_{1}$

Step 2.3

Cancel the common factor of $4$ and $10$ .

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Step 2.3.1

Factor $2$ out of $4$ .

$\int \frac{2 (2)}{10 e^{u_{1}}} d u_{1}$

Step 2.3.2

Cancel the common factors.

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Step 2.3.2.1

Factor $2$ out of $10 e^{u_{1}}$ .

$\int \frac{2 (2)}{2 (5 e^{u_{1}})} d u_{1}$

Step 2.3.2.2

Cancel the common factor.

$\int \frac{2 \cdot 2}{2 (5 e^{u_{1}})} d u_{1}$

Step 2.3.2.3

Rewrite the expression.

$\int \frac{2}{5 e^{u_{1}}} d u_{1}$

Step 3

Since $\frac{2}{5}$ is constant with respect to $u_{1}$ , move $\frac{2}{5}$ out of the integral.

$\frac{2}{5} \int \frac{1}{e^{u_{1}}} d u_{1}$

Step 4

Simplify the expression.

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Step 4.1

Negate the exponent of $e^{u_{1}}$ and move it out of the denominator.

$\frac{2}{5} \int 1 {(e^{u_{1}})}^{- 1} d u_{1}$

Step 4.2

Simplify.

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Step 4.2.1

Multiply the exponents in ${(e^{u_{1}})}^{- 1}$ .

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Step 4.2.1.1

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

$\frac{2}{5} \int 1 e^{u_{1} \cdot - 1} d u_{1}$

Step 4.2.1.2

Move $- 1$ to the left of $u_{1}$ .

$\frac{2}{5} \int 1 e^{- 1 \cdot u_{1}} d u_{1}$

Step 4.2.1.3

Rewrite $- 1 u_{1}$ as $- u_{1}$ .

$\frac{2}{5} \int 1 e^{- u_{1}} d u_{1}$

Step 4.2.2

Multiply $e^{- u_{1}}$ by $1$ .

$\frac{2}{5} \int e^{- u_{1}} d u_{1}$

Step 5

Let $u_{2} = - u_{1}$ . Then $d u_{2} = - d u_{1}$ , so $- d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 5.1

Let $u_{2} = - u_{1}$ . Find $\frac{d u_{2}}{d u_{1}}$ .

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Step 5.1.1

Differentiate $- u_{1}$ .

$\frac{d}{d u_{1}} [- u_{1}]$

Step 5.1.2

Since $- 1$ is constant with respect to $u_{1}$ , the derivative of $- u_{1}$ with respect to $u_{1}$ is $- \frac{d}{d u_{1}} [u_{1}]$ .

$- \frac{d}{d u_{1}} [u_{1}]$

Step 5.1.3

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 1$ .

$- 1 \cdot 1$

Step 5.1.4

Multiply $- 1$ by $1$ .

$- 1$

Step 5.2

Rewrite the problem using $u_{2}$ and $d u_{2}$ .

$\frac{2}{5} \int - e^{u_{2}} d u_{2}$

Step 6

Since $- 1$ is constant with respect to $u_{2}$ , move $- 1$ out of the integral.

$\frac{2}{5} (- \int e^{u_{2}} d u_{2})$

Step 7

The integral of $e^{u_{2}}$ with respect to $u_{2}$ is $e^{u_{2}}$ .

$\frac{2}{5} (- (e^{u_{2}} + C))$

Step 8

Simplify.

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Step 8.1

Simplify.

$\frac{2}{5} (- e^{u_{2}}) + C$

Step 8.2

Combine $\frac{2}{5}$ and $e^{u_{2}}$ .

$- \frac{2 e^{u_{2}}}{5} + C$

Step 9

Substitute back in for each integration substitution variable.

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Step 9.1

Replace all occurrences of $u_{2}$ with $- u_{1}$ .

$- \frac{2 e^{- u_{1}}}{5} + C$

Step 9.2

Replace all occurrences of $u_{1}$ with $5 x^{2}$ .

$- \frac{2 e^{- (5 x^{2})}}{5} + C$

Step 10

Multiply $5$ by $- 1$ .

$- \frac{2 e^{- 5 x^{2}}}{5} + C$

Step 11

Reorder terms.

$- \frac{2}{5} e^{- 5 x^{2}} + C$