Integral Calculator: Solve Definite and Indefinite Integrals (Antiderivatives)

Online integral calculator with step-by-step solutions, interactive graphs, and image scanning. Choose solution steps manually. Methods include u-substitution, integration by parts, partial fractions, trigonometric/hyperbolic/logarithmic formulas, polynomial long division, Euler substitution, binomial differentials, Weierstrass substitution, and Ostrogradsky's method. Solve definite and improper integrals using the Fundamental Theorem of Calculus.

Integration variable

Upper limit			+∞
Lower limit			-∞

Copy solution link

75% 90% 100% 110% 125% 🔍

Check derivative of antiderivative

Original integral

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

Antiderivative

\frac{ln ( ∣ x + 1 ∣ )}{12} - \frac{ln ( x ^{2} - x + 1 )}{6} + \frac{arctan ( \frac{2 x - 1}{3} )}{3 3} + \frac{1}{6 ( x + 1 )} + \frac{ln ( ∣ x - 1 ∣ )}{4} + C

Step-by-step solution

Calculating

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

Factoring the denominator

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

Partial Fraction Decomposition

Decomposing the rational function into partial fractions

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

Bringing the terms on the right-hand side to a common denominator

\frac{x ^{2}}{( x - 1 ) ( x + 1 ) ^{2} ( x ^{2} - x + 1 )} = \frac{( A + C + E ) x ^{4} + ( A + B - C + D + E ) x ^{3} + ( - A + B - 2 D ) x ^{2} + ( - A - B + C + 2 D + E ) x + ( - B - C - D + E )}{( x - 1 ) ( x + 1 ) ^{2} ( x ^{2} - x + 1 )}

Equating coefficients of like powers of

x^{n}

in the numerators on both sides

x^{0} : x^{1} : x^{2} : x^{3} : x^{4} : ⎩ ⎨ ⎧ - B - C - D + E = 0 - A - B + C + 2 D + E = 0 - A + B - 2 D = 1 A + B - C + D + E = 0 A + C + E = 0 = ⎩ ⎨ ⎧ A = - \frac{1}{3} B = \frac{1}{3} C = \frac{1}{12} D = - \frac{1}{6} E = \frac{1}{4}

Substituting the found coefficients into the partial fractions form

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

\int \frac{\frac{1}{3} - \frac{x}{3}}{x ^{2} - x + 1} + \frac{1}{12 ( x + 1 )} - \frac{1}{6 ( x + 1 ) ^{2}} + \frac{1}{4 ( x - 1 )} d x

Rewriting the expression

\int - \frac{x - 1}{3 ( x ^{2} - x + 1 )} + \frac{1}{12 ( x + 1 )} - \frac{1}{6 ( x + 1 ) ^{2}} + \frac{1}{4 ( x - 1 )} d x

Linearity

Details

1. Sum/Difference rule:

• The integral of a sum or difference equals the sum or difference of the integrals:

$\int [f (x) + g (x)] d x = \int f (x) d x + \int g (x) d x$

$\int [f (x) - g (x)] d x = \int f (x) d x - \int g (x) d x$

• Example:

$\int (2 x + 3) d x = \int 2 x d x + \int 3 d x$

2. Constant multiple rule:

• Constants can be factored out of the integral:

$\int a \cdot f (x) d x = a \cdot \int f (x) d x$

• Example:

$\int 5 x d x = 5 \int x d x$

3. Linearity (combined property):

• Integrals are linear operators, meaning they satisfy both rules above:

$\int [a \cdot f (x) + b \cdot g (x)] d x =$

$= a \cdot \int f (x) d x + b \cdot \int g (x) d x$

• Example:

$\int (2 x^{2} + 3 sin (x)) d x =$

$= 2 \int x^{2} d x + 3 \int sin (x) d x$

- \frac{1}{3} \int \frac{x - 1}{x ^{2} - x + 1} d x + \frac{1}{12} \int \frac{1}{x + 1} d x - \frac{1}{6} \int \frac{1}{( x + 1 ) ^{2}} d x + \frac{1}{4} \int \frac{1}{x - 1} d x =

= - \frac{1}{3} \cdot (1) + \frac{1}{12} \cdot (4) - \frac{1}{6} \cdot (5) + \frac{1}{4} \cdot (6)

= \frac{ln ( ∣ x + 1 ∣ )}{12} - \frac{ln ( x ^{2} - x + 1 )}{6} + \frac{arctan ( \frac{2 x - 1}{3} )}{3 3} + \frac{1}{6 ( x + 1 )} + \frac{ln ( ∣ x - 1 ∣ )}{4}

Calculating

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

(1)

Rewriting the integral using the identity

x - 1 = \frac{1}{2} (2 x - 1) - \frac{1}{2}

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

= \frac{1}{2} \cdot (2) + (3)

= \frac{ln ( x ^{2} - x + 1 )}{2} - \frac{arctan ( \frac{2 x - 1}{3} )}{3}

Calculating

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

(2)

u-Substitution

u = x^{2} - x + 1 d u = (2 x - 1) d x

[1]

\int \frac{1}{u} d u

Table integral

Details

Table of integrals:

1. $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ where $n \neq = - 1$

2. $\int \frac{1}{x} d x = ln ∣ x ∣ + C$ where $x \neq = 0$

3. $\int \frac{1}{x ^{2} + 1} d x = {arctan (x) + C - arccot (x) + C$

4. $\int \frac{1}{1 - x ^{2}} d x = \frac{1}{2} ln ∣ ∣ \frac{1 + x}{1 - x} ∣ ∣ + C$

5. $\int \frac{1}{1 - x ^{2}} d x = {arcsin (x) + C - arccos (x) + C$

6. $\int \frac{1}{x ^{2} \pm 1} d x = ln ∣ ∣ x + x^{2} \pm 1 ∣ ∣ + C$

7. $\int e^{x} d x = e^{x} + C$

8. $\int a^{x} d x = \frac{a ^{x}}{ln ( a )} + C$ where $a > 0$ , $a \neq = 1$

9. $\int sin (x) d x = - cos (x) + C$

10. $\int cos (x) d x = sin (x) + C$

11. $\int \frac{1}{sin ^{2} ( x )} d x = - cot (x) + C$

12. $\int \frac{1}{cos ^{2} ( x )} d x = tan (x) + C$

13. $\int sinh (x) d x = cosh (x) + C$

14. $\int cosh (x) d x = sinh (x) + C$

15. $\int \frac{1}{sinh ^{2} ( x )} d x = - coth (x) + C$

16. $\int \frac{1}{cosh ^{2} ( x )} d x = tanh (x) + C$

ln (u)

Back-substitution

u = x^{2} - x + 1

[1]

ln (x^{2} - x + 1)

Calculating

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

(3)

Using completing the square

Details

How to complete the square

For the quadratic expression $a x^{2} + b x + c = 0$ :

1. Factor out $a$ (if $a \neq = 1$ ):

$a x^{2} + b x + c = a (x^{2} + \frac{b}{a} x) + c$

2. Add and subtract $(\frac{b}{2 a})^{2}$ inside the parentheses:

$a (x^{2} + \frac{b}{a} x + (\frac{b}{2 a})^{2} - (\frac{b}{2 a})^{2}) + c$

3. Rewrite as a perfect square trinomial:

$a ((x + \frac{b}{2 a})^{2} - (\frac{b}{2 a})^{2}) + c$

4. Simplify the constants:

$a (x + \frac{b}{2 a})^{2} - \frac{b ^{2}}{4 a} + c$

Example: Complete the square for $x^{2} + 4 x - 5$ .

1. Add/subtract $(\frac{4}{2})^{2} = 4$ :

$x^{2} + 4 x + 4 - 4 - 5$

2. Form the perfect square:

$(x + 2)^{2} - 9$

- \frac{1}{2} \int \frac{1}{( x - \frac{1}{2} ) ^{2} + \frac{3}{4}} d x

u-Substitution

v = x - \frac{1}{2} x = \frac{2 v + 1}{2} d x = d v

[2]

- \frac{1}{2} \int \frac{1}{v ^{2} + \frac{3}{4}} d v

u-Substitution

t = \frac{2 v}{3} v = \frac{3 t}{2} d v = \frac{3}{2} d t

[3]

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

Factoring out the common factor

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

Table integral

- \frac{1}{3} \cdot arctan (t) = - \frac{arctan ( t )}{3}

Back-substitution

t = \frac{2 v}{3} v = \frac{3 t}{2}

[3]

- \frac{arctan ( \frac{2 v}{3} )}{3}

Back-substitution

v = x - \frac{1}{2} x = \frac{2 v + 1}{2}

[2]

- \frac{arctan ( \frac{2 x - 1}{3} )}{3}

Calculating

\int \frac{1}{x + 1} d x

(4)

Rewriting with differentials

\int \frac{1}{x + 1} d (x + 1)

Table integral

\int \frac{1}{u} d u = ln (∣ u ∣)

ln (∣ x + 1 ∣)

Calculating

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

(5)

u-Substitution

u = x + 1 x = u - 1 d x = d u

[4]

\int \frac{1}{u ^{2}} d u

Using the power rule

\int u^{n} d u = \frac{u ^{n + 1}}{n + 1}, (n \neq = - 1)

where

n = - 2

- \frac{1}{u}

Back-substitution

u = x + 1 x = u - 1

[4]

- \frac{1}{x + 1}

Calculating

\int \frac{1}{x - 1} d x

(6)

Rewriting with differentials

\int \frac{1}{x - 1} d (x - 1)

Table integral

\int \frac{1}{u} d u = ln (∣ u ∣)

ln (∣ x - 1 ∣)

ln (x)

arctan (x)

d x

C

Bookmark this page — CTRL+D