Calculus Examples

$\int \frac{x^{2}}{{(x - 1)}^{4}} d x$

Step 1

Let $u_{1} = x - 1$ . Then $d u_{1} = d x$ . Rewrite using $u_{1}$ and $d$ $u_{1}$ .

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Step 1.1

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

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Step 1.1.1

Differentiate $x - 1$ .

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

Step 1.1.2

By the Sum Rule, the derivative of $x - 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 1]$

Step 1.1.3

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

$1 + \frac{d}{d x} [- 1]$

Step 1.1.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$1 + 0$

Step 1.1.5

Add $1$ and $0$ .

$1$

Step 1.2

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

$\int \frac{{(u_{1} + 1)}^{2}}{{u_{1}}^{4}} d u_{1}$

Step 2

Apply basic rules of exponents.

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Step 2.1

Move ${u_{1}}^{4}$ out of the denominator by raising it to the $- 1$ power.

$\int {(u_{1} + 1)}^{2} {({u_{1}}^{4})}^{- 1} d u_{1}$

Step 2.2

Multiply the exponents in ${({u_{1}}^{4})}^{- 1}$ .

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Step 2.2.1

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

$\int {(u_{1} + 1)}^{2} {u_{1}}^{4 \cdot - 1} d u_{1}$

Step 2.2.2

Multiply $4$ by $- 1$ .

$\int {(u_{1} + 1)}^{2} {u_{1}}^{- 4} d u_{1}$

Step 3

Let $u_{2} = u_{1} + 1$ . Then $d u_{2} = d u_{1}$ . Rewrite using $u_{2}$ and $d$ $u_{2}$ .

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Step 3.1

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

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Step 3.1.1

Differentiate $u_{1} + 1$ .

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

Step 3.1.2

By the Sum Rule, the derivative of $u_{1} + 1$ with respect to $u_{1}$ is $\frac{d}{d u_{1}} [u_{1}] + \frac{d}{d u_{1}} [1]$ .

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

Step 3.1.3

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

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

Step 3.1.4

Since $1$ is constant with respect to $u_{1}$ , the derivative of $1$ with respect to $u_{1}$ is $0$ .

$1 + 0$

Step 3.1.5

Add $1$ and $0$ .

$1$

Step 3.2

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

$\int {u_{2}}^{2} {(u_{2} - 1)}^{- 4} d u_{2}$

Step 4

Let $u_{3} = u_{2} - 1$ . Then $d u_{3} = d u_{2}$ . Rewrite using $u_{3}$ and $d$ $u_{3}$ .

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Step 4.1

Let $u_{3} = u_{2} - 1$ . Find $\frac{d u_{3}}{d u_{2}}$ .

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Step 4.1.1

Differentiate $u_{2} - 1$ .

$\frac{d}{d u_{2}} [u_{2} - 1]$

Step 4.1.2

By the Sum Rule, the derivative of $u_{2} - 1$ with respect to $u_{2}$ is $\frac{d}{d u_{2}} [u_{2}] + \frac{d}{d u_{2}} [- 1]$ .

$\frac{d}{d u_{2}} [u_{2}] + \frac{d}{d u_{2}} [- 1]$

Step 4.1.3

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

$1 + \frac{d}{d u_{2}} [- 1]$

Step 4.1.4

Since $- 1$ is constant with respect to $u_{2}$ , the derivative of $- 1$ with respect to $u_{2}$ is $0$ .

$1 + 0$

Step 4.1.5

Add $1$ and $0$ .

$1$

Step 4.2

Rewrite the problem using $u_{3}$ and $d u_{3}$ .

$\int {(u_{3} + 1)}^{2} {u_{3}}^{- 4} d u_{3}$

Step 5

Simplify.

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Step 5.1

Rewrite ${(u_{3} + 1)}^{2}$ as $(u_{3} + 1) (u_{3} + 1)$ .

$\int (u_{3} + 1) (u_{3} + 1) {u_{3}}^{- 4} d u_{3}$

Step 5.2