Question

Calculus 2 Master needed, show steps with partial fraction decomposition B26%29%2F%28%28x-5%29%28x%5E2%2B4%29%29%7D%20%5C%2C%20dx" id="TexFormula1" title="\int(3x^2-26x+26)/((x-5)(x^2+4))} \, dx" alt="\int(3x^2-26x+26)/((x-5)(x^2+4))} \, dx" align="absmiddle" class="latex-formula">

Answer 1

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.

$\frac{3x^{2}-26x+26}{(x-5)(x^{2}+4)}=\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}$

Combine the two fractions back into one using the common denominator.

$\frac{A}{x-5} +\frac{Bx+C}{x^{2}+4}=\frac{A(x^{2}+4)+(Bx+C)(x-5)}{(x-5)(x^{2}+4)}$

This numerator will equal the original numerator.

$A(x^{2}+4)+(Bx+C)(x-5)=3x^{2}-26x+26\\Ax^{2}+4A+Bx^{2}-5Bx+Cx-5C=3x^{2}-26x+26\\(A+B)x^{2}+(C-5B)x+(4A-5C)=3x^{2}-26x+26$

Match the coefficients.

$A+B=3\\C-5B=-26\\4A-5C=26$

Solve the system of equations.

$A=-1\\B=4\\C=-6$

So we can rewrite the integral as:

$\int {(\frac{-1}{x-5}+\frac{4x-6}{x^{2}+4}) } \, dx$

Solving:

$\int {\frac{-1}{x-5}\, dx + \int {\frac{4x}{x^{2}+4}} \, dx - \int {\frac{6}{x^{2}+4} } \, dx$

$-\int {\frac{1}{x-5}\, dx + 2\int {\frac{2x}{x^{2}+4}} \, dx - 6\int {\frac{1}{x^{2}+4} } \, dx$

$-ln|x-5| + 2ln(x^{2}+4) - 6(\frac{1}{2} tan^{-1}(\frac{x}{2} )) + C$

$-ln|x-5| + 2ln(x^{2}+4) - 3 tan^{-1}(\frac{x}{2} ) + C$