Question

Calculus 2 Master needed, show steps with partial fraction decomposition
`\int(3x^2-26x+26)/((x-5)(x^2+4))} \, dx`

Answer 1

-ln|x−5| + 2 ln(x²+4) + 3 tan⁻¹(x/2) + C

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.

`(3x^(2)-26x+26)/((x-5)(x^(2)+4))=(A)/(x-5) +(Bx+C)/(x^(2)+4)`

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

`(A)/(x-5) +(Bx+C)/(x^(2)+4)=(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 {((-1)/(x-5)+(4x-6)/(x^(2)+4)) } \, dx`

Solving:

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

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

`-ln|x-5| + 2ln(x^(2)+4) - 6((1)/(2) tan^(-1)((x)/(2) )) + C`

`-ln|x-5| + 2ln(x^(2)+4) - 3 tan^(-1)((x)/(2) ) + C`