Answer:
-ln|x−5| + 2 ln(x²+4) + 3 tan⁻¹(x/2) + C
Step-by-step explanation:
The fraction will be split into a sum of two other fractions.
The first fraction will have a denominator of x − 5. The numerator will the a polynomial of one less order, in this case, a constant A.
The second fraction will have a denominator of x² + 4. The numerator will be Bx + C.
[tex]\frac{3x^{2}-26x+26}{(x-5)(x^{2}+4)}=\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}[/tex]
Combine the two fractions back into one using the common denominator.
[tex]\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}=\frac{A(x^{2}+4)+(Bx+C)(x-5)}{(x-5)(x^{2}+4)}[/tex]
This numerator will equal the original numerator.
[tex]A(x^{2}+4)+(Bx+C)(x-5)=3x^{2}-26x+26\\Ax^{2}+4A+Bx^{2}-5Bx+Cx-5C=3x^{2}-26x+26\\(A+B)x^{2}+(C-5B)x+(4A-5C)=3x^{2}-26x+26[/tex]
Match the coefficients.
[tex]A+B=3\\C-5B=-26\\4A-5C=26[/tex]
Solve the system of equations.
[tex]A=-1\\B=4\\C=-6[/tex]
So we can rewrite the integral as:
[tex]\int {(\frac{-1}{x-5}+\frac{4x-6}{x^{2}+4}) } \, dx[/tex]
Solving:
[tex]\int {\frac{-1}{x-5}\, dx + \int {\frac{4x}{x^{2}+4}} \, dx - \int {\frac{6}{x^{2}+4} } \, dx[/tex]
[tex]-\int {\frac{1}{x-5}\, dx + 2\int {\frac{2x}{x^{2}+4}} \, dx - 6\int {\frac{1}{x^{2}+4} } \, dx[/tex]
[tex]-ln|x-5| + 2ln(x^{2}+4) - 6(\frac{1}{2} tan^{-1}(\frac{x}{2} )) + C[/tex]
[tex]-ln|x-5| + 2ln(x^{2}+4) - 3 tan^{-1}(\frac{x}{2} ) + C[/tex]
