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
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