How do you integrate $\frac{x^{3} + 25}{x^{2} + 4 x + 3}$ $(x^3+25)/(x^2+4x+3)$ using partial fractions?

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How do you integrate $\frac{x^{3} + 25}{x^{2} + 4 x + 3}$ $(x^3+25)/(x^2+4x+3)$ using partial fractions?

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Answer 1 · 2017-07-08T19:16:47

$12 ln | x + 1 | + ln | x + 3 | + c$ $12ln|x+1|+ln|x+3|+c$

Explanation:

$factorising the numerator$ $"factorising the numerator"$

$\frac{x^{3} + 25}{(x + 1) (x + 3)}$ $(x^3+25)/((x+1)(x+3)$

$\Rightarrow \frac{x^{3} + 25}{(x + 1) (x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 3}$ $rArr(x^3+25)/((x+1)(x+3))=A/(x+1)+B/(x+3)$

$multiply through by (x + 1) (x + 3)$ $"multiply through by " (x+1)(x+3)$

$\Rightarrow x^{3} + 25 = A (x + 3) + B (x + 1)$ $rArrx^3+25=A(x+3)+B(x+1)$

$using the cover up method$ $"using the "color(blue)"cover up method"$

$x = - 3 \to - 2 = - 2 B \Rightarrow B = 1$ $x=-3to-2=-2BrArrB=1$

$x = - 1 \to 24 = 2 A \Rightarrow A = 12$ $x=-1to24=2ArArrA=12$

$\Rightarrow \int \frac{x^{3} + 25}{x^{2} + 4 x + 3} d x = \int \frac{12}{x + 1} d x + \int \frac{1}{x + 3} d x$ $rArrint(x^3+25)/(x^2+4x+3)dx=int12/(x+1)dx+int1/(x+3)dx$

$= 12 ln | x + 1 | + ln | x + 3 | + c$ $=12ln|x+1|+ln|x+3|+c$

Answer 2 · 2017-07-09T16:12:42

$\int \frac{x^{3} + 25}{x^{2} + 4 x + 3} d x$ $int(x^3+25)/(x^2+4x+3)dx$

$= \int (x - 4) d x + \int \frac{13 x + 27}{x^{2} + 4 x + 3} d x$ $=int(x-4)dx+int(13x+27)/(x^2+4x+3)dx$

$= \frac{x^{2}}{2} - 4 x + \int \frac{13 x + 27}{(x + 1) \cdot (x + 3)} d x$ $=x^2/2-4x+int(13x+27)/((x+1)*(x+3))dx$

Now I decomposed $\frac{13 x + 27}{(x + 1) \cdot (x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 3}$ $(13x+27)/((x+1)*(x+3))=A/(x+1)+B/(x+3)$ fraction

$A \cdot (x + 3) + B \cdot (x + 1) = 13 x + 27$ $A*(x+3)+B*(x+1)=13x+27$

$(A + B) \cdot x + 3 A + B = 13 x + 27$ $(A+B)*x+3A+B=13x+27$

After equating coefficients, I found $A + B = 13$ $A+B=13$ and $3 A + B = 27$ $3A+B=27$ equations,

After solving them simultaneously, $A = 7$ $A=7$ and $B = 6$ $B=6$

Thus,

$\int \frac{x^{3} + 25}{x^{2} + 4 x + 3} d x$ $int(x^3+25)/(x^2+4x+3)dx$

$= \frac{x^{2}}{2} - 4 x + \int \frac{13 x + 27}{(x + 1) \cdot (x + 3)} d x$ $=x^2/2-4x+int(13x+27)/((x+1)*(x+3))dx$

$= \frac{x^{2}}{2} - 4 x + \int \frac{7}{x + 1} d x + \int \frac{6}{x + 3} d x$ $=x^2/2-4x+int7/(x+1)dx+int6/(x+3)dx$

$= \frac{x^{2}}{2} - 4 x + 7 ln (x + 1) + 6 ln (x + 3) + C$ $=x^2/2-4x+7Ln(x+1)+6Ln(x+3)+C$

Explanation:

1) I took long division

2) I decomposed second integral into basic fractions

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How do you integrate $\frac{x^{3} + 25}{x^{2} + 4 x + 3}$ $(x^3+25)/(x^2+4x+3)$ using partial fractions?

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