How do you integrate $\int \frac{x + 1}{(6 x^{2} + 4) (x - 5)}$ $int(x+1)/((6x^2+4)(x-5))$ using partial fractions?

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How do you integrate $\int \frac{x + 1}{(6 x^{2} + 4) (x - 5)}$ $int(x+1)/((6x^2+4)(x-5))$ using partial fractions?

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Answer 1 · 2017-10-02T17:41:00

See below.

Explanation:

1) I used basic fractions method.

Inıtıially, I decomposed integrand into basic fractions,

$\frac{x + 1}{(6 x^{2} + 4) \cdot (x - 5)} = \frac{A}{x - 5} + \frac{B x + C}{6 x^{2} + 4}$ $(x+1)/[(6x^2+4)*(x-5)]=A/(x-5)+(Bx+C)/(6x^2+4)$

After expanding denominator,

$A \cdot (6 x^{2} + 4) + (B x + C) \cdot (x - 5) = x + 1$ $A*(6x^2+4)+(Bx+C)*(x-5)=x+1$

$(6 A + B) \cdot x^{2} + (C - 5 B) \cdot x + 4 A - 5 C = x + 1$ $(6A+B)*x^2+(C-5B)*x+4A-5C=x+1$

After equating coefficients, I found

$6 A + B = 0$ $6A+B=0$ , $C - 5 B = 1$ $C-5B=1$ and $4 A - 5 C = 1$ $4A-5C=1$ equations.

After solving them, $A = \frac{3}{77}, B = - \frac{18}{77} and C = - \frac{13}{77}$ $A=3/77, B=-18/77 and C=-13/77$

Thus,

$\int \frac{(x + 1) d x}{(6 x^{2} + 4) \cdot (x - 5)}$ $int ((x+1)dx)/[(6x^2+4)*(x-5)]$

= $\frac{3}{77}$ $3/77$ $\int \frac{d x}{x - 5}$ $int dx/(x-5)$ - $\frac{1}{77}$ $1/77$ $\int \frac{(18 x + 13) \cdot d x}{6 x^{2} + 4}$ $int ((18x+13)*dx)/(6x^2+4)$

= $\frac{3}{77} \cdot ln (x - 5)$ $3/77*ln(x-5)$ - $\frac{3}{154}$ $3/154$ $\int \frac{12 x \cdot d x}{6 x^{2} + 4}$ $int (12x*dx)/(6x^2+4)$ - $\frac{13}{77}$ $13/77$ $\int \frac{d x}{6 x^{2} + 4}$ $int dx/(6x^2+4)$

= $\frac{3}{77} \cdot ln (x - 5) - \frac{3}{154} \cdot ln (6 x^{2} + 4) - \frac{13 \sqrt{6}}{924} \cdot arctan (\frac{3 x}{\sqrt{6}}) + C$ $3/77*ln(x-5)-3/154*ln(6x^2+4)-(13sqrt(6))/924*arctan((3x)/sqrt(6))+C$

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How do you integrate $\int \frac{x + 1}{(6 x^{2} + 4) (x - 5)}$ $int(x+1)/((6x^2+4)(x-5))$ using partial fractions?

1 Answer

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How do you integrate $\int \frac{x + 1}{(6 x^{2} + 4) (x - 5)}$ $int(x+1)/((6x^2+4)(x-5))$ using partial fractions?