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Integration by Partial Fractions

Evaluate:∫ 3x...

Question

$E v a l u a t e : \int \frac{(3 x + 5) d x}{\sqrt{x^{2} + 4 x + 3}} .$

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Solution

$\int \frac{(3 x + 5) d x}{\sqrt{(} x^{2} + 4 x + 3)}$
If u= $x^{2} + 4 x + 3$ .
$d u = (2 x + 4) d x$
$3 \int \frac{x + \frac{5}{3}}{\sqrt{(} x^{2} + 4 x + 3)}$
= $\frac{3}{2} \int \frac{2 x + \frac{10}{3}}{\sqrt{(} x^{2} + 4 x + 3)}$
= $\frac{3}{2} \int \frac{2 x + 4 - 4 + \frac{10}{3}}{\sqrt{(} x^{2} + 4 x + 3)} d x$
= $\frac{3}{2} \int \frac{2 x + 4}{\sqrt{(} x^{2} + 4 x + 3)} d x - \frac{3}{2} \int \frac{\frac{2}{3}}{\sqrt{(} x^{2} + 4 x + 3)}$
= $\frac{3}{2} \int \frac{d u}{\sqrt{(} u)} - \int \frac{1}{\sqrt{(} u)}$
$I_{1}$ = $\frac{3}{2} \int (u)^{\frac{- 1}{2}} d u$
$I_{1}$ = $\frac{3}{2} \frac{\sqrt{(} u)}{\frac{1}{2}} + C_{1}$
= $3 \sqrt{(} u) + C_{1}$
Hence, $I_{1} = 3 \sqrt{(} x^{2} + 4 x + 3) + C_{1}$
$I_{2} = - \int \frac{1}{\sqrt{(} x^{2} + 4 x + 3)} d x$
= $- \int \frac{d x}{\sqrt{(} x^{2} + 4 x + 4 - 4 + 3)}$
= $- \int \frac{1}{\sqrt{(} (x + 2)^{2} - (1)^{2}} d x$
= $\int \frac{d x}{\sqrt{(} x^{2} - a^{2})}$ = $log ∣ x + \sqrt{(} x^{2} - a^{2}) ∣ + C_{2} = log ∣ x + 2 + \sqrt{(} x + 2)^{2} - 1) ∣ + C_{2}$
I= $3 \sqrt{(} x^{2} + 4 x + 3) - log ∣ x + 2 + \sqrt{(} x^{2} + 4 x + 3) ∣ + C$

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