evaluate the indefinite integral.\n int \frac { 7 ( x ^ { 2 } + 4 ) } { x ( x ^ { 2 } + 7 ) } d x = square…

evaluate the indefinite integral.\n int \frac { 7 ( x ^ { 2 } + 4 ) } { x ( x ^ { 2 } + 7 ) } d x = square \n(use parentheses to clearly denote the argument of each function.)

evaluate the indefinite integral.\n int \frac { 7 ( x ^ { 2 } + 4 ) } { x ( x ^ { 2 } + 7 ) } d x = square \n(use parentheses to clearly denote the argument of each function.)

Answer

Explanation:

Step1: Perform partial fraction decomposition

Let (\frac{7(x^{2}+4)}{x(x^{2}+7)}=\frac{A}{x}+\frac{Bx + C}{x^{2}+7}). Then (7(x^{2}+4)=A(x^{2}+7)+(Bx + C)x=(A + B)x^{2}+Cx+7A). Equating coefficients:

  • For (x^{2}): (7=A + B)
  • For (x): (0 = C)
  • For the constant term: (28 = 7A), so (A = 4). Since (A = 4) and (7=A + B), then (B=3) and (C = 0). So (\frac{7(x^{2}+4)}{x(x^{2}+7)}=\frac{4}{x}+\frac{3x}{x^{2}+7}).

Step2: Integrate term - by - term

(\int\frac{7(x^{2}+4)}{x(x^{2}+7)}dx=\int(\frac{4}{x}+\frac{3x}{x^{2}+7})dx) Using the integral formulas (\int\frac{1}{u}du=\ln|u|+C) ((u=x) for the first term) and (\int\frac{u}{a^{2}+u^{2}}du=\frac{1}{2}\ln|a^{2}+u^{2}|+C) ((u = x^{2}+7), (du = 2xdx), so (\int\frac{3x}{x^{2}+7}dx=\frac{3}{2}\int\frac{2x}{x^{2}+7}dx)): (\int\frac{4}{x}dx+\int\frac{3x}{x^{2}+7}dx=4\ln|x|+\frac{3}{2}\ln(x^{2}+7)+C)

Answer:

(4\ln|x|+\frac{3}{2}\ln(x^{2}+7)+C)