Calculus Examples

$\int \frac{1}{x^{3} - 2 x^{2} + x} d x d x$

Step 1

Write the fraction using partial fraction decomposition.

Tap for more steps...

Step 1.1

Decompose the fraction and multiply through by the common denominator.

Tap for more steps...

Step 1.1.1

Factor the fraction.

Tap for more steps...

Step 1.1.1.1

Factor $x$ out of $x^{3} - 2 x^{2} + x$ .

Tap for more steps...

Step 1.1.1.1.1

Factor $x$ out of $x^{3}$ .

$\frac{1}{x \cdot x^{2} - 2 x^{2} + x}$

Step 1.1.1.1.2

Factor $x$ out of $- 2 x^{2}$ .

$\frac{1}{x \cdot x^{2} + x (- 2 x) + x}$

Step 1.1.1.1.3

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

$\frac{1}{x \cdot x^{2} + x (- 2 x) + x^{1}}$

Step 1.1.1.1.4

Factor $x$ out of $x^{1}$ .

$\frac{1}{x \cdot x^{2} + x (- 2 x) + x \cdot 1}$

Step 1.1.1.1.5

Factor $x$ out of $x \cdot x^{2} + x (- 2 x)$ .

$\frac{1}{x (x^{2} - 2 x) + x \cdot 1}$

Step 1.1.1.1.6

Factor $x$ out of $x (x^{2} - 2 x) + x \cdot 1$ .

$\frac{1}{x (x^{2} - 2 x + 1)}$

Step 1.1.1.2

Factor using the perfect square rule.

Tap for more steps...

Step 1.1.1.2.1

Rewrite $1$ as $1^{2}$ .

$\frac{1}{x (x^{2} - 2 x + 1^{2})}$

Step 1.1.1.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 1.1.1.2.3

Rewrite the polynomial.

$\frac{1}{x (x^{2} - 2 \cdot x \cdot 1 + 1^{2})}$

Step 1.1.1.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

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

Step 1.1.2

For each factor in the denominator, create a new fraction using the factor as the denominator, and an unknown value as the numerator. Since the factor in the denominator is linear, put a single variable in its place $B$ .

$\frac{A}{x} + \frac{B}{{(x - 1)}^{2}}$

Step 1.1.3

$\frac{A}{x} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{x - 1}$

Step 1.1.4

Multiply each fraction in the equation by the denominator of the original expression. In this case, the denominator is $x {(x - 1)}^{2}$ .

$\frac{1 (x {(x - 1)}^{2})}{x {(x - 1)}^{2}} = \frac{A (x {(x - 1)}^{2})}{x} + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.5

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.1.5.1

Cancel the common factor.

$\frac{1 (x {(x - 1)}^{2})}{x {(x - 1)}^{2}} = \frac{A (x {(x - 1)}^{2})}{x} + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.5.2

Rewrite the expression.

$\frac{1 ({(x - 1)}^{2})}{{(x - 1)}^{2}} = \frac{A (x {(x - 1)}^{2})}{x} + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.6

Cancel the common factor of ${(x - 1)}^{2}$ .

Tap for more steps...

Step 1.1.6.1

Cancel the common factor.

$\frac{1 {(x - 1)}^{2}}{{(x - 1)}^{2}} = \frac{A (x {(x - 1)}^{2})}{x} + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.6.2

Rewrite the expression.

$1 = \frac{A (x {(x - 1)}^{2})}{x} + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7

Simplify each term.

Tap for more steps...

Step 1.1.7.1

Cancel the common factor of $x$ .

Tap for more steps...

Step 1.1.7.1.1

Cancel the common factor.

$1 = \frac{A (x {(x - 1)}^{2})}{x} + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.1.2

Divide $A ({(x - 1)}^{2})$ by $1$ .

$1 = A ({(x - 1)}^{2}) + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.2

Rewrite ${(x - 1)}^{2}$ as $(x - 1) (x - 1)$ .

$1 = A ((x - 1) (x - 1)) + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.3

Expand $(x - 1) (x - 1)$ using the FOIL Method.

Tap for more steps...

Step 1.1.7.3.1

Apply the distributive property.

$1 = A (x (x - 1) - 1 (x - 1)) + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.3.2

Apply the distributive property.

$1 = A (x \cdot x + x \cdot - 1 - 1 (x - 1)) + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.3.3

Apply the distributive property.

$1 = A (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1) + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.4

Simplify and combine like terms.

Tap for more steps...

Step 1.1.7.4.1

Simplify each term.

Tap for more steps...

Step 1.1.7.4.1.1

Multiply $x$ by $x$ .

$1 = A (x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1) + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.4.1.2

Move $- 1$ to the left of $x$ .

$1 = A (x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1) + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.4.1.3

Rewrite $- 1 x$ as $- x$ .

$1 = A (x^{2} - x - 1 x - 1 \cdot - 1) + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.4.1.4

Rewrite $- 1 x$ as $- x$ .

$1 = A (x^{2} - x - x - 1 \cdot - 1) + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.4.1.5

Multiply $- 1$ by $- 1$ .

$1 = A (x^{2} - x - x + 1) + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.4.2

Subtract $x$ from $- x$ .

$1 = A (x^{2} - 2 x + 1) + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.5

Apply the distributive property.

$1 = A x^{2} + A (- 2 x) + A \cdot 1 + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.6

Simplify.

Tap for more steps...

Step 1.1.7.6.1

Rewrite using the commutative property of multiplication.

$1 = A x^{2} - 2 A x + A \cdot 1 + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.6.2

Multiply $A$ by $1$ .

$1 = A x^{2} - 2 A x + A + \frac{(B) (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.7

Cancel the common factor of ${(x - 1)}^{2}$ .

Tap for more steps...

Step 1.1.7.7.1

Cancel the common factor.

$1 = A x^{2} - 2 A x + A + \frac{B (x {(x - 1)}^{2})}{{(x - 1)}^{2}} + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.7.2

Divide $(B) (x)$ by $1$ .

$1 = A x^{2} - 2 A x + A + (B) (x) + \frac{(C) (x {(x - 1)}^{2})}{x - 1}$

Step 1.1.7.8

Cancel the common factor of ${(x - 1)}^{2}$ and $x - 1$ .

Tap for more steps...

Step 1.1.7.8.1

Factor $x - 1$ out of $(C) (x {(x - 1)}^{2})$ .

$1 = A x^{2} - 2 A x + A + B x + \frac{(x - 1) (C (x (x - 1)))}{x - 1}$

Step 1.1.7.8.2

Cancel the common factors.

Tap for more steps...

Step 1.1.7.8.2.1

Multiply by $1$ .

$1 = A x^{2} - 2 A x + A + B x + \frac{(x - 1) (C (x (x - 1)))}{(x - 1) \cdot 1}$

Step 1.1.7.8.2.2

Cancel the common factor.

$1 = A x^{2} - 2 A x + A + B x + \frac{(x - 1) (C (x (x - 1)))}{(x - 1) \cdot 1}$

Step 1.1.7.8.2.3

Rewrite the expression.

$1 = A x^{2} - 2 A x + A + B x + \frac{C (x (x - 1))}{1}$

Step 1.1.7.8.2.4

Divide $C (x (x - 1))$ by $1$ .

$1 = A x^{2} - 2 A x + A + B x + C (x (x - 1))$

Step 1.1.7.9

Apply the distributive property.

$1 = A x^{2} - 2 A x + A + B x + C (x \cdot x + x \cdot - 1)$

Step 1.1.7.10

Multiply $x$ by $x$ .

$1 = A x^{2} - 2 A x + A + B x + C (x^{2} + x \cdot - 1)$

Step 1.1.7.11

Move $- 1$ to the left of $x$ .

$1 = A x^{2} - 2 A x + A + B x + C (x^{2} - 1 \cdot x)$

Step 1.1.7.12

Rewrite $- 1 x$ as $- x$ .

$1 = A x^{2} - 2 A x + A + B x + C (x^{2} - x)$

Step 1.1.7.13

Apply the distributive property.

$1 = A x^{2} - 2 A x + A + B x + C x^{2} + C (- x)$

Step 1.1.7.14

Rewrite using the commutative property of multiplication.

$1 = A x^{2} - 2 A x + A + B x + C x^{2} - C x$

Step 1.1.8

Simplify the expression.

Tap for more steps...

Step 1.1.8.1

Move $A$ .

$1 = A x^{2} - 2 x A + A + B x + C x^{2} - C x$

Step 1.1.8.2

Reorder $B$ and $x$ .

$1 = A x^{2} - 2 x A + A + B x + C x^{2} - C x$

Step 1.1.8.3

Move $A$ .

$1 = A x^{2} - 2 x A + B x + C x^{2} - C x + A$

Step 1.1.8.4

Move $B x$ .

$1 = A x^{2} - 2 x A + C x^{2} + B x - C x + A$

Step 1.1.8.5

Move $- 2 x A$ .

$1 = A x^{2} + C x^{2} - 2 x A + B x - C x + A$

Step 1.2

Create equations for the partial fraction variables and use them to set up a system of equations.

Tap for more steps...

Step 1.2.1

Create an equation for the partial fraction variables by equating the coefficients of $x^{2}$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = A + C$

Step 1.2.2

Create an equation for the partial fraction variables by equating the coefficients of $x$ from each side of the equation. For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$0 = - 2 A + B - 1 C$

Step 1.2.3

Create an equation for the partial fraction variables by equating the coefficients of the terms not containing $x$ . For the equation to be equal, the equivalent coefficients on each side of the equation must be equal.

$1 = A$

Step 1.2.4

Set up the system of equations to find the coefficients of the partial fractions.

$0 = A + C$

$0 = - 2 A + B - 1 C$

$1 = A$

$0 = A + C$

$0 = - 2 A + B - 1 C$

$1 = A$

Step 1.3

Solve the system of equations.

Tap for more steps...

Step 1.3.1

Rewrite the equation as $A = 1$ .

$A = 1$

$0 = A + C$

$0 = - 2 A + B - 1 C$

Step 1.3.2

Replace all occurrences of $A$ with $1$ in each equation.

Tap for more steps...

Step 1.3.2.1

Replace all occurrences of $A$ in $0 = A + C$ with $1$ .

$0 = (1) + C$

$A = 1$

$0 = - 2 A + B - 1 C$

Step 1.3.2.2

Simplify the right side.

Tap for more steps...

Step 1.3.2.2.1

Remove parentheses.

$0 = 1 + C$

$A = 1$

$0 = - 2 A + B - 1 C$

$0 = 1 + C$

$A = 1$

$0 = - 2 A + B - 1 C$

Step 1.3.2.3

Replace all occurrences of $A$ in $0 = - 2 A + B - 1 C$ with $1$ .

$0 = - 2 \cdot 1 + B - 1 C$

$0 = 1 + C$

$A = 1$

Step 1.3.2.4

Simplify the right side.

Tap for more steps...

Step 1.3.2.4.1

Simplify each term.

Tap for more steps...

Step 1.3.2.4.1.1

Multiply $- 2$ by $1$ .

$0 = - 2 + B - 1 C$

$0 = 1 + C$

$A = 1$

Step 1.3.2.4.1.2

Rewrite $- 1 C$ as $- C$ .

$0 = - 2 + B - C$

$0 = 1 + C$

$A = 1$

$0 = - 2 + B - C$

$0 = 1 + C$

$A = 1$

$0 = - 2 + B - C$

$0 = 1 + C$

$A = 1$

$0 = - 2 + B - C$

$0 = 1 + C$

$A = 1$

Step 1.3.3

Solve for $C$ in $0 = 1 + C$ .

Tap for more steps...

Step 1.3.3.1

Rewrite the equation as $1 + C = 0$ .

$1 + C = 0$

$0 = - 2 + B - C$

$A = 1$

Step 1.3.3.2

Subtract $1$ from both sides of the equation.

$C = - 1$

$0 = - 2 + B - C$

$A = 1$

$C = - 1$

$0 = - 2 + B - C$

$A = 1$

Step 1.3.4

Replace all occurrences of $C$ with $- 1$ in each equation.

Tap for more steps...

Step 1.3.4.1

Replace all occurrences of $C$ in $0 = - 2 + B - C$ with $- 1$ .

$0 = - 2 + B - (- 1)$

$C = - 1$

$A = 1$

Step 1.3.4.2

Simplify the right side.

Tap for more steps...

Step 1.3.4.2.1

Simplify $- 2 + B - (- 1)$ .

Tap for more steps...

Step 1.3.4.2.1.1

Multiply $- 1$ by $- 1$ .

$0 = - 2 + B + 1$

$C = - 1$

$A = 1$

Step 1.3.4.2.1.2

Add $- 2$ and $1$ .

$0 = B - 1$

$C = - 1$

$A = 1$

$0 = B - 1$

$C = - 1$

$A = 1$

$0 = B - 1$

$C = - 1$

$A = 1$

$0 = B - 1$

$C = - 1$

$A = 1$

Step 1.3.5

Solve for $B$ in $0 = B - 1$ .

Tap for more steps...

Step 1.3.5.1

Rewrite the equation as $B - 1 = 0$ .

$B - 1 = 0$

$C = - 1$

$A = 1$

Step 1.3.5.2

Add $1$ to both sides of the equation.

$B = 1$

$C = - 1$

$A = 1$

$B = 1$

$C = - 1$

$A = 1$

Step 1.3.6

Solve the system of equations.

$B = 1 C = - 1 A = 1$

Step 1.3.7

List all of the solutions.

$B = 1, C = - 1, A = 1$

Step 1.4

Replace each of the partial fraction coefficients in $\frac{A}{x} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{x - 1}$ with the values found for $A$ , $B$ , and $C$ .

$\frac{1}{x} + \frac{1}{{(x - 1)}^{2}} + \frac{- 1}{x - 1}$

Step 1.5

Move the negative in front of the fraction.

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

Step 2

Split the single integral into multiple integrals.

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

Step 3

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

$(\ln (| x |) + C + \int \frac{1}{{(x - 1)}^{2}} d x + \int - \frac{1}{x - 1} d x) d x$

Step 4

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

Tap for more steps...

Step 4.1

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

Tap for more steps...

Step 4.1.1

Differentiate $x - 1$ .

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

Step 4.1.2

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 4.1.3

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 4.1.4

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

$1 + 0$

Step 4.1.5

Add $1$ and $0$ .

$1$

Step 4.2

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

$(\ln (| x |) + C + \int \frac{1}{{u_{1}}^{2}} d u_{1} + \int - \frac{1}{x - 1} d x) d x$

Step 5

Apply basic rules of exponents.

Tap for more steps...

Step 5.1

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

$(\ln (| x |) + C + \int {({u_{1}}^{2})}^{- 1} d u_{1} + \int - \frac{1}{x - 1} d x) d x$

Step 5.2

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

Tap for more steps...

Step 5.2.1

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

$(\ln (| x |) + C + \int {u_{1}}^{2 \cdot - 1} d u_{1} + \int - \frac{1}{x - 1} d x) d x$

Step 5.2.2

Multiply $2$ by $- 1$ .

$(\ln (| x |) + C + \int {u_{1}}^{- 2} d u_{1} + \int - \frac{1}{x - 1} d x) d x$

Step 6

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

$(\ln (| x |) + C - {u_{1}}^{- 1} + C + \int - \frac{1}{x - 1} d x) d x$

Step 7

Since $- 1$ is constant with respect to $x$ , move $- 1$ out of the integral.

$(\ln (| x |) + C - {u_{1}}^{- 1} + C - \int \frac{1}{x - 1} d x) d x$

Step 8

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

Tap for more steps...

Step 8.1

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

Tap for more steps...

Step 8.1.1

Differentiate $x - 1$ .

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

Step 8.1.2

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 8.1.3

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 8.1.4

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

$1 + 0$

Step 8.1.5

Add $1$ and $0$ .

$1$

Step 8.2

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

$(\ln (| x |) + C - {u_{1}}^{- 1} + C - \int \frac{1}{u_{2}} d u_{2}) d x$

Step 9

The integral of $\frac{1}{u_{2}}$ with respect to $u_{2}$ is $\ln (| u_{2} |)$ .

$(\ln (| x |) + C - {u_{1}}^{- 1} + C - (\ln (| u_{2} |) + C)) d x$

Step 10

Simplify.

$(- {u_{1}}^{- 1} + \ln (\frac{| x |}{| u_{2} |})) d x + C$

Step 11

Substitute back in for each integration substitution variable.

Tap for more steps...

Step 11.1

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

$(- {(x - 1)}^{- 1} + \ln (\frac{| x |}{| u_{2} |})) d x + C$

Step 11.2

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

$(- {(x - 1)}^{- 1} + \ln (\frac{| x |}{| x - 1 |})) d x + C$