Evaluate the integral. (Use C for the constant of integration.) 7x3 5x2 49x 5 (x2 1)(x2 7) dx

[tex]\int{\frac{7x^3+5x^2+49x+5}{(x^2+1)(x^2+7)} } \, dx\\ \\\frac{7x^3+5x^2+49x+5}{(x^2+1)(x^2+7)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+7}\\ \\7x^3+5x^2+49x+5=(x^2+7)(Ax+B)+(x^2+1)(Cx+D)\\\\7x^3+5x^2+49x+5=Ax^3+Bx^2+7Ax+7B+Cx^3+Dx^2+Cx+D\\\\7x^3+5x^2+49x+5=Ax^3+Cx^3+Bx^2+Dx^2+7Ax+Cx+7B+D\\\\7x^3+5x^2+49x+5=x^3(A+C)+x^2(B+D)+x(7A+C)+7B+D[/tex]

Set up a system of equations and solve for each constant

[tex]\begin{cases} A + C = 7\\B + D = 5\\7 A + C = 49\\7 B + D = 5 \end{cases}[/tex]

[tex]A+C=7\\A=7-C[/tex]

[tex]7A+C=49\\7(7-C)+C=49\\49-7C+C=49\\49-6C=49\\-6C=0\\C=0[/tex]

[tex]A=7-C\\A=7-0\\A=7[/tex]

[tex]B+D=5\\B=5-D[/tex]

[tex]7B+D=5\\7(5-D)+D=5\\35-7D+D=5\\35-6D=5\\-6D=-30\\D=5[/tex]

[tex]B=5-D\\B=5-5\\B=0[/tex]

Plug solved constants in and evaluate

[tex]\frac{7x^3+5x^2+49x+5}{(x^2+1)(x^2+7)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+7}\\\\\frac{7x^3+5x^2+49x+5}{(x^2+1)(x^2+7)}=\frac{7x+0}{x^2+1}+\frac{0x+5}{x^2+7}\\\\\frac{7x^3+5x^2+49x+5}{(x^2+1)(x^2+7)}=\frac{7x}{x^2+1}+\frac{5}{x^2+7}[/tex]

Break up integral

[tex]\int {\bigr(\frac{7x}{x^2+1}+\frac{5}{x^2+7}\bigr) } \, dx[/tex]

[tex]\int {\frac{7x}{x^2+1} } \, dx +\int {\frac{5}{x^2+7}} \, dx\\\\\int {\frac{7x}{x^2+1} } \, dx +5\int {\frac{1}{x^2+7}} \, dx[/tex]

Solve first integral

Let [tex]u=x^2+1[/tex] and [tex]du=2xdx[/tex] for the first integral. Thus, [tex]\frac{7}{2}du=7xdx[/tex]:

[tex]\frac{7}{2}\int {\frac{du}{u}}\\\\\frac{7}{2}ln(|u|)+C\\ \\\frac{7}{2}ln(|x^2+1|)+C[/tex]

Solve second integral

Since [tex]5\int {\frac{1}{x^2+7} } \, dx[/tex] is in the form of [tex]\int{\frac{1}{x^2+a^2} } \, dx[/tex], its formula is [tex]\frac{1}{a}arctan(\frac{x}{a})+C[/tex]:

[tex]5\int {\frac{1}{x^2+7} } \, dx\\\\5\bigr(\frac{1}{\sqrt{7}}arctan(\frac{x}{\sqrt{7}})+C\bigr)\\\\\frac{5}{\sqrt{7}}arctan(\frac{x}{\sqrt{7}})+C[/tex]

Combine integrals

[tex]\biggr[\frac{7}{2}ln(|x^2+1|)+C\biggr]+\biggr[\frac{5}{\sqrt{7}}arctan(\frac{x}{\sqrt{7}})+C\biggr]\\ \\\frac{7}{2}ln(|x^2+1|)+\frac{5}{\sqrt{7}}arctan(\frac{x}{\sqrt{7}})+C[/tex]