Question

Evaluate the integral. (Use C for the constant of integration. Enter your answer using function notation - use ln(x) instead of ln x.)

Answer 1

Answer:

$\int\limits {\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} \, dx = \frac{3}{2}\ln(x^2 + 1) + 4\sqrt{3}\tan^{-1}(\frac{x}{\sqrt 3} )+ c$

Step-by-step explanation:

Given

$\int\limits {\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} \, dx$

Required

Integrate

Using partial fraction, we have:

$\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} = \frac{Ax+B}{x^2 + 1} + \frac{Cx + D}{x^2 + 3}$

Take LCM

$\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} = \frac{(Ax+B)(x^2 + 3)+ (Cx + D)(x^2 + 1)}{(x^2 + 1)(x^2 + 3)}$

Cancel out the denominators

$3x^3 + 4x^2 + 9x + 4} = (Ax+B)(x^2 + 3)+ (Cx + D)(x^2 + 1)$

Open brackets

$3x^3 + 4x^2 + 9x + 4} = Ax^3+Bx^2 + 3Ax +3B+ Cx^3 + Dx^2 + Cx + D$

Collect like terms

$3x^3 + 4x^2 + 9x + 4 = Ax^3+ Cx^3+Bx^2+ Dx^2 + 3Ax+ Cx +3B + D$

Compare like terms on opposite sides

$Ax^3 + Cx^3 = 3x^3$              $A + C = 3$

$Bx^2 + Dx^2 = 4x^2$              $B + D = 4$

$3Ax + Cx = 9x$                  $3A + C = 9$

$3B + D = 4$

Subtract $B + D = 4$ from $3B + D = 4$

$3B - B + D - D = 4 - 4$

$2B + 0 = 0$

$2B = 0$

$B = 0$

$B + D = 4$

$D =4 - B$

$D =4 - 0$

$D =4$

Subtract $A + C = 3$ from $3A + C = 9$

$3A - A + C - C = 9 - 3$

$2A = 6$

$A = 3$

$A + C = 3$

$C = 3 - A$

$C = 3 - 3$

$C = 0$

So, we have:

$A = 3$        $B = 0$       $C = 0$       $D =4$

$\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} = \frac{Ax+B}{x^2 + 1} + \frac{Cx + D}{x^2 + 3}$

$\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} = \frac{3x+0}{x^2 + 1} + \frac{0*x + 4}{x^2 + 3}$

$\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} = \frac{3x}{x^2 + 1} + \frac{4}{x^2 + 3}$

The integral becomes:

$\int\limits {[\frac{3x}{x^2 + 1} + \frac{4}{x^2 + 3}]} \, dx$

Split:

$\int\limits {\frac{3x}{x^2 + 1} \, dx + \int\limits {\frac{4}{x^2 + 3}} \, dx$

Split

$\frac{3}{2} \int\limits {\frac{2x}{x^2 + 1} \, dx + 4\int\limits {\frac{1}{x^2 + 3}} \ dx$

Integrate

$\frac{3}{2}\ln(x^2 + 1) + 4\sqrt{3}\tan^{-1}\frac{x}{\sqrt 3} + c$

Hence:

$\int\limits {\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} \, dx = \frac{3}{2}\ln(x^2 + 1) + 4\sqrt{3}\tan^{-1}(\frac{x}{\sqrt 3} )+ c$