I = ∫3x+4/x^2+6x+5dx = ∫3x+9 5/(x^2+6x+5)dx ⇒∫(2x+1)dx/x^2+6x+5 5∫dx/(x^2+6x+5) (A) ∫(3x+9)dx/x^2+6x+5 = ∫3(x+3)/(x^2+6x+5)dx #160; (x^2+6 ...

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Integration by Partial Fractions

I=∫3x+4/x2+6x...

Question

$I = \int \frac{3 x + 4}{x^{2} + 6 x + 5} d x$

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Solution

$I = \int \frac{3 x + 4}{x^{2} + 6 x + 5} d x = \int \frac{3 x + 9 - 5}{(x^{2} + 6 x + 5)} d x$
$\Rightarrow \int \frac{(2 x + 1) d x}{x^{2} + 6 x + 5} - 5 \int \frac{d x}{(x^{2} + 6 x + 5)}$
$(A) \int \frac{(3 x + 9) d x}{x^{2} + 6 x + 5} = \int \frac{3 (x + 3)}{(x^{2} + 6 x + 5)} d x$
$(x^{2} + 6 x + 5) = P$
$(2 x + 6) d x = d p$
$2 (x + 3) d x = d p$
$A \Rightarrow \int \frac{3 d p}{2 (p)} \Rightarrow \frac{3}{2} l n | p | + c \Rightarrow \frac{3}{2} l n ∣ ∣ x^{2} + 6 x + 5 ∣ ∣ + c$
$(B) \to \int \frac{5 d x}{(x^{2} + 6 x + 5)} = \int \frac{5 d x}{(x + 3)^{2} - 4} \Rightarrow \int \frac{5 d x}{(x + 3)^{2} - (2)^{2}}$
$x + 3 = p$
$d x = d p$
$B \Rightarrow 5 \int \frac{d p}{p^{2} - 2^{2}}$
$\Rightarrow 5 \int d p [\frac{1}{(p + 2) (p - 2)}] \Rightarrow \frac{5}{4} \int d p [\frac{1}{(p - 2)} - \frac{1}{(p + 2)}]$
$\Rightarrow \frac{5}{4} l n | p - 2 | - \frac{5}{4} l n | p + 2 | \Rightarrow \frac{5}{4} l n ∣ ∣ \frac{p - 2}{p + 2} ∣ ∣$
$∴ I = A - B = \frac{3}{2} l n ∣ ∣ x^{2} + 6 x + 5 ∣ ∣ - \frac{5}{4} l n ∣ ∣ \frac{p - 2}{p + 2} ∣ ∣ + c$
$∴ \int \frac{3 x + 4}{x^{2} + 6 x + 5} d x = \frac{3}{2} l n ∣ ∣ x^{2} + 6 x + 5 ∣ ∣ - \frac{5}{4} l n ∣ ∣ \frac{p - 2}{p + 2} ∣ ∣ + c$
$= \frac{3}{2} l n ∣ ∣ x^{2} + 6 x + 5 ∣ ∣ - \frac{5}{4} ∣ ∣ \frac{x + 1}{x + 5} ∣ ∣ + c$

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