Mathematics

# Evaluate: $\int \dfrac {x + 2}{\sqrt {x^{2} + 5x + 6}} dx.$

##### SOLUTION
Let $I=\int \dfrac{x+2}{\sqrt{x^2+5x+6}}dx$
Put $x + 2=\lambda\left ( \dfrac{d}{dx}(x^2+5x+6) \right )+\mu$
$x+2=2\lambda x+5\lambda+\mu$
Comparing coefficients of x both sides
$1 = 2\lambda\Rightarrow \lambda=\dfrac{1}{2}$
Comparing constant terms both sides,
$2=5\lambda+\mu$
or, $2=5(\dfrac{1}{2})+\mu$
$\mu=2-\dfrac{5}{2}=\dfrac{-1}{2}$
Therefore, $\int \dfrac{x+2}{\sqrt{x^2+5x+6}}dx=\int \dfrac{\dfrac{1}{2}(2x+5)-\dfrac{1}{2}}{\sqrt{x^2+5x+6}}dx$    $(x+2=\lambda(2x+5)+\mu)$
$I=\int \dfrac{\dfrac{1}{2}(2x+5)}{\sqrt{x^2+5x+6}}dx-\dfrac{1}{2}\int\dfrac{dx}{\sqrt{x^2+5x+6}}$
Therefore, $I=I_1-I_2$---- (1)
$I_1= \int \dfrac{\dfrac{1}{2}(2x+5)}{\sqrt{x^2+5x+6}}dx$
Put $x^2+5x+6=t$
$\therefore (2x+5)dx=dt$
$=\dfrac{1}{2}\int \dfrac{dt}{\sqrt{t}}=\dfrac{1}{2}\left ( \dfrac{t^{\dfrac{-1}{2}+1}}{\dfrac{-1}{2}+1} \right )+C$
$=t^{\frac{1}{2}}+C=\sqrt{t}+C\Rightarrow \sqrt{x^2+5x+6}+C$
$I_2=\dfrac{1}{2}\int\dfrac{dx}{\sqrt{x^2+5x+6}}$
$\dfrac{1}{2}\int\dfrac{dx}{\sqrt{x^2+5x+\dfrac{25}{4}-\dfrac{25}{4}+6}}=\dfrac{1}{2}\int \dfrac{dx}{\sqrt{\left ( x+\dfrac{5}{2} \right )^2-\left ( \dfrac{1}{2} \right )^2}}$
$\dfrac{1}{2}.log\left [ \left ( x+\dfrac{5}{2} \right )+\sqrt{\left ( x+\dfrac{5}{2} \right )^2-\left ( \dfrac{1}{2} \right )^2} \right ]+C$
$\dfrac{1}{2}.log\left [ \left ( x+\dfrac{5}{2} \right )+\sqrt{x^2+5x+6} \right ]+C$
Substituting the values, of $I_1$ and $I_2$ in (1),
$I=\sqrt{x^2+5x+6}+\dfrac{1}{2}.log\left [ \left ( x+\dfrac{5}{2} \right )+\sqrt{x^2+5x+6} \right ]+C$

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Subjective Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 124

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