How do you integrate $\int \frac{3 x^{2} - 2 x + 5}{{(x^{2} + 1)}^{2}}$ $int (3x^2-2x+5)/(x^2+1)^2$ using partial fractions?

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How do you integrate $\int \frac{3 x^{2} - 2 x + 5}{{(x^{2} + 1)}^{2}}$ $int (3x^2-2x+5)/(x^2+1)^2$ using partial fractions?

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Answer 1 · 2017-06-11T21:41:24

$\int \frac{3 x^{2} - 2 x + 5}{{(x^{2} + 1)}^{2}} d x = 4 arctan x + \frac{x + 1}{x^{2} + 1} + C$ $int (3x^2-2x+5)/(x^2+1)^2 dx = 4arctanx + (x+1)/(x^2+1) +C$

Explanation:

Write the numerator as:

$3 x^{2} - 2 x + 5 = 3 (x^{2} + 1) - 2 x + 2$ $3x^2-2x+5 = 3(x^2+1) -2x +2$

so that:

$\int \frac{3 x^{2} - 2 x + 5}{{(x^{2} + 1)}^{2}} d x = 3 \int \frac{d x}{1 + x^{2}} - \int \frac{2 x d x}{{(x^{2} + 1)}^{2}} + 2 \int \frac{d x}{{(x^{2} + 1)}^{2}}$ $int (3x^2-2x+5)/(x^2+1)^2 dx = 3int dx/(1+x^2) - int (2xdx)/(x^2+1)^2 +2 int dx/(x^2+1)^2$

Now solve the integral separately:

$\int \frac{d x}{1 + x^{2}} = arctan x + C_{1}$ $int dx/(1+x^2) = arctanx + C_1$

$\int \frac{2 x d x}{{(x^{2} + 1)}^{2}} = \int \frac{d (x^{2} + 1)}{{(x^{2} + 1)}^{2}} = - \frac{1}{x^{2} + 1} + C_{2}$ $int (2xdx)/(x^2+1)^2 = int (d(x^2+1))/(x^2+1)^2 = -1/(x^2+1)+C_2$

The third integral can be resolved by substituting:

$x = tan t$ $x=tant$

$d x = \frac{d t}{{cos}^{2} t}$ $dx = dt/cos^2t$

$\int \frac{d x}{{(x^{2} + 1)}^{2}} = \int \frac{d t}{{cos}^{2} t {(1 + {tan}^{2} t)}^{2}}$ $int dx/(x^2+1)^2 = int dt/(cos^2t(1+tan^2t)^2)$

use now the identity:

$1 + {tan}^{2} t = \frac{1}{{cos}^{2} t}$ $1+tan^2t = 1/cos^2t$

$\int \frac{d x}{{(x^{2} + 1)}^{2}} = \int \frac{d t}{\frac{{cos}^{2} t}{{cos}^{4} t}} = \int {cos}^{2} t d t$ $int dx/(x^2+1)^2 = int dt/(cos^2t/cos^4t) = int cos^2tdt$

and as ${cos}^{2} t = \frac{1 + cos 2 t}{2}$ $cos^2t = (1+cos2t)/2$

$\int \frac{d x}{{(x^{2} + 1)}^{2}} = \frac{1}{2} \int d t + \frac{1}{2} \int cos (2 t) d t$ $int dx/(x^2+1)^2 = 1/2 int dt + 1/2 int cos(2t)dt$

$\int \frac{d x}{{(x^{2} + 1)}^{2}} = \frac{t}{2} + \frac{1}{4} \int cos (2 t) d (2 t)$ $int dx/(x^2+1)^2 = t/2 + 1/4 int cos(2t)d(2t)$

$\int \frac{d x}{{(x^{2} + 1)}^{2}} = \frac{t}{2} + \frac{sin (2 t)}{4} + C_{3}$ $int dx/(x^2+1)^2 = t/2 + sin(2t)/4+C_3$

Use now the parametric formula:

$sin (2 t) = \frac{2 tan t}{1 + {tan}^{2} t}$ $sin(2t) = (2tant)/(1+tan^2t)$

to undo the substitution:

$\int \frac{d x}{{(x^{2} + 1)}^{2}} = \frac{arctan x}{2} + \frac{1}{2} \frac{x}{1 + x^{2}} + C_{3}$ $int dx/(x^2+1)^2 = arctanx/2 + 1/2x/(1+x^2)+C_3$

Putting it together:

$\int \frac{3 x^{2} - 2 x + 5}{{(x^{2} + 1)}^{2}} d x = 3 arctan x + \frac{1}{x^{2} + 1} + 2 (\frac{arctan x}{2} + \frac{1}{2} \frac{x}{1 + x^{2}}) + C$ $int (3x^2-2x+5)/(x^2+1)^2 dx = 3arctanx +1/(x^2+1) + 2(arctanx/2 + 1/2x/(1+x^2))+C$

$\int \frac{3 x^{2} - 2 x + 5}{{(x^{2} + 1)}^{2}} d x = 4 arctan x + \frac{x + 1}{x^{2} + 1} + C$ $int (3x^2-2x+5)/(x^2+1)^2 dx = 4arctanx + (x+1)/(x^2+1) +C$

Answer 2 · 2017-06-12T00:07:15

This answer will only look at the partial fraction decomposition of the integrand as its integral has already been evaluated in the other answer.

Explanation:

$\frac{3 x^{2} - 2 x + 5}{{(x^{2} + 1)}^{2}} \equiv \frac{A + B x}{x^{2} + 1} + \frac{C + D x}{{(x^{2} + 1)}^{2}}$ $(3x^2-2x+5)/(x^2+1)^2-=(A+Bx)/(x^2+1)+(C+Dx)/(x^2+1)^2$

$3 x^{2} - 2 x + 5 = (A + B x) (x^{2} + 1) + C + D x$ $3x^2-2x+5=(A+Bx)(x^2+1)+C+Dx$

$3 x^{2} - 2 x + 5 = A + A x^{2} + B x^{3} + (B + D) x + C$ $3x^2-2x+5=A+Ax^2+Bx^3+(B+D)x+C$

Compare coefficients of all $x$ $x$ -terms

$5 = A + C$ $5=A+C$ (1)

$- 2 = B + D$ $-2=B+D$ (2)

$3 = A$ $3=A$ (3)

$0 = B$ $0=B$ (4)

Sub (4) into (2) and (3) into (1)

$- 2 = D$ $-2=D$

$5 = 3 + C \Rightarrow C = 2$ $5=3+C rArr C=2$

$\frac{3 x^{2} - 2 x + 5}{{(x^{2} + 1)}^{2}} = \frac{3}{x^{2} + 1} + \frac{2 - 2 x}{{(x^{2} + 1)}^{2}}$ $(3x^2-2x+5)/(x^2+1)^2=3/(x^2+1)+(2-2x)/(x^2+1)^2$

This can be rewritten as:

$\frac{3}{x^{2} + 1} - \frac{2 (x - 1)}{{(x^{2} + 1)}^{2}}$ $3/(x^2+1)-(2(x-1))/(x^2+1)^2$

$\int \frac{3}{x^{2} + 1} - \frac{2 (x - 1)}{{(x^{2} + 1)}^{2}}$ $int3/(x^2+1)-(2(x-1))/(x^2+1)^2$ $d x = 4 arctan x + \frac{x + 1}{x^{2} + 1} + c$ $dx= 4arctanx + (x+1)/(x^2+1) +"c"$

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How do you integrate $\int \frac{3 x^{2} - 2 x + 5}{{(x^{2} + 1)}^{2}}$ $int (3x^2-2x+5)/(x^2+1)^2$ using partial fractions?

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