integral of (7x^2+9x+7)/((x^2+1)^2)

1) Let's integrate:

\int \frac{7 x^{2} + 9 x + 7}{{(x^{2} + 1)}^{2}} d x

ⓘ A square being never negative,

x^{2} + 1

is never zero and cannot be factored in ℝ

2) You can decompose into partial fractions:

\frac{7 x^{2} + 9 x + 7}{{(x^{2} + 1)}^{2}} = \frac{A + B x}{x^{2} + 1} + \frac{C + D x}{{(x^{2} + 1)}^{2}}

To find the remaining unknowns, you multiply both sides by the denominator:

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

which yields:

7 x^{2} + 9 x + 7 = (A + B x) \cdot (x^{2} + 1) + (C + D x)

7 x^{2} + 9 x + 7 = (A + B x) \cdot (x^{2} + 1) + C + D x

ⓘ Expand by multiplying each term inside the parentheses:

(A + B x) \cdot (x^{2} + 1)

A x^{2}

+ A \cdot 1

+ B x x^{2}

+ B x \cdot 1

7 x^{2} + 9 x + 7 = A x^{2}

+ A \cdot 1

+ B x x^{2}

+ B x \cdot 1

+ C

+ D x

ⓘ Multiplying powers with the same base adds the exponents:

x x^{2}

x^{3}

7 x^{2} + 9 x + 7 = A x^{2}

+ A

+ B x^{3}

+ B x

+ C

+ D x

3) By identifying the different terms, you obtain the system of equations:

[1]:

0 = B

[2]:

7 = A

[3]:

9 = B + D

[4]:

7 = A + C

Solution obtained [1]:

B = 0

4) Let's solve for A:

7 = A

[2]

Solution found [2]:

A = 7

5) Solving for D:

9 = B + D

[3] using

B = 0

ⓘ Adding 0 changes nothing:

0 + D

D

Solution obtained [3]:

D = 9

6) Let's solve for C:

7 = A + C

[4] where

A = 7

7 + C = 7

ⓘ We subtract

7

from both sides

C = 7 - 7

ⓘ Start from

7

and move back

7

on the number line, down to zero:

7 - 7

0

Solution found [4]:

C = 0

7) Finally, substitute the unknowns back into the initial partial fraction decomposition:

\frac{7 x^{2} + 9 x + 7}{{(x^{2} + 1)}^{2}} = \frac{7}{x^{2} + 1} + \frac{9 x}{{(x^{2} + 1)}^{2}}

8) Let's integrate:

\int (\frac{9 x}{{(x^{2} + 1)}^{2}} + \frac{7}{x^{2} + 1}) d x

For the math expression:

\frac{7}{x^{2} + 1}

, since (arctan(a))' = a'/(a^2 + 1), we try differentiating

\arctan (x)

9) Let's perform the differentiation:

\frac{d}{d x} (\arctan (x))

ⓘ The derivative of arctan(x) is 1/(1 + x^2):

\frac{d}{d x} (\arctan (x))

\frac{1}{x^{2} + 1}

\frac{1}{x^{2} + 1}

The computed derivative matches the original math expression up to a constant factor. Therefore, the integral is:

ⅰ

\int (\frac{9 x}{{(x^{2} + 1)}^{2}} + \frac{7}{x^{2} + 1}) d x

7 \cdot \arctan (x) + \int \frac{9 x}{{(x^{2} + 1)}^{2}} d x

7 \cdot \arctan (x) + \int \frac{9 x}{{(x^{2} + 1)}^{2}} d x

For the math expression:

\frac{9 x}{{(x^{2} + 1)}^{2}}

, since (a^n)' = n*a'*a^(n - 1), we try differentiating

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

10) Let's perform the differentiation:

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

ⓘ The exponent is computed before the power:

- 2 + 1

-1

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

Apply the differentiation rule:

\frac{d}{d x} (f^{n}) = n \cdot \frac{d}{d x} (f) \cdot f^{n - 1}

with f =

x^{2} + 1

ⓘ The derivative of x^n is n*x^(n-1):

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

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

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

ⓘ Multiplying by -1 gives the opposite:

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

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

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

ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

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

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

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

ⓘ The derivative of a sum is the sum of the derivatives:

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

\frac{d}{d x} (x^{2}) + 0

- \frac{\frac{d}{d x} (x^{2}) + 0}{{(x^{2} + 1)}^{2}}

ⓘ Adding 0 changes nothing:

\frac{d}{d x} (x^{2}) + 0

\frac{d}{d x} (x^{2})

■The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (x^{2})

2 x

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

The computed derivative matches the original math expression up to a constant factor. Therefore, the integral is:

ⓘ The derivative of

- \frac{9}{2} \cdot \frac{1}{x^{2} + 1}

\frac{9 x}{{(x^{2} + 1)}^{2}}

, so

- \frac{9}{2} \cdot \frac{1}{x^{2} + 1}

is an antiderivative of

\frac{9 x}{{(x^{2} + 1)}^{2}}

\int \frac{9 x}{{(x^{2} + 1)}^{2}} d x

- \frac{9}{2} \cdot \frac{1}{x^{2} + 1}

7 \cdot \arctan (x) - \frac{9}{2} \cdot \frac{1}{x^{2} + 1}

ⓘ To multiply fractions, multiply the numerators together and the denominators together:

\frac{9}{2} \cdot \frac{1}{x^{2} + 1}

\frac{9 \cdot 1}{2 \cdot (x^{2} + 1)}

\frac{9}{2 \cdot (x^{2} + 1)}

ⅰ

\int \frac{7 x^{2} + 9 x + 7}{{(x^{2} + 1)}^{2}} d x

7 \cdot \arctan (x) - \frac{9}{2 \cdot (x^{2} + 1)}

7 \cdot \arctan (x) - \frac{9}{2 \cdot (x^{2} + 1)}

▶Indefinite integrals are defined up to an additive constant C, so this yields:

7 \cdot \arctan (x) - \frac{9}{2 \cdot (x^{2} + 1)} + C