Mathematics

Evaluate the following integrals:
$$\int { \sqrt { 2{ x }^{ 2 }+3x+4 }  } dx$$


SOLUTION
Let $$I=\displaystyle\int \sqrt{2 x^2+3 x+4} d x$$

        $$=\displaystyle\int \sqrt{2\left(x^2+2(x)\bigg(\dfrac{3}{4}\bigg)+2\right)}$$

        $$=\sqrt{2}\displaystyle\int \sqrt{x^2+2(x)\bigg(\dfrac{3}{4}\bigg)+\bigg(\dfrac{3}{4}\bigg)^2-\bigg(\dfrac{3}{4}\bigg)^2+2}\ d x$$

       $$=\sqrt{2}\displaystyle\int \sqrt{\bigg(x+\dfrac{3}{4}\bigg)^2+\dfrac{23}{16}}\ d x$$                                                        $$(\because (a+b)^2=a^2+2 a b+b^2)$$

Put $$t=x+\dfrac{3}{4}\implies d t=d x$$

$$I=\sqrt{2}\displaystyle\int \sqrt{t^2+\bigg(\dfrac{\sqrt{23}}{4}\bigg)^2}\ d t$$

As we know that

$$\displaystyle\int \sqrt{a^2+x^2} d x=\dfrac{x}{2}\sqrt{x^2+a^2}+\dfrac{a^2}{2}\ln |x+\sqrt{x^2+a^2}|+C$$

Here $$a=\dfrac{\sqrt{23}}{4}$$

$$\implies I=\sqrt{2}\left(\dfrac{t}{2}\sqrt{t^2+\dfrac{23}{16}}+\dfrac{23/16}{2}\ln \left|t+\sqrt{t^2+\dfrac{23}{16}}\right|+C\right)$$

$$\implies I=\dfrac{1}{2}\bigg(x+\dfrac{3}{4}\bigg)\sqrt{2\bigg(x+\dfrac{3}{4}\bigg)^2+\dfrac{23}{16}}+\dfrac{23\sqrt{2}}{32}\ln \left|x+\dfrac{3}{4}+\sqrt{\bigg(x+\dfrac{3}{4}\bigg)^2+\dfrac{23}{16}}\ \right|+C$$

$$\implies I=\dfrac{1}{8}(4  x+3)\sqrt{2 x^2+3 x+4}+\dfrac{23\sqrt{2}}{32}\ln \left|x+\dfrac{3}{4}+\sqrt{x^2+\dfrac{3}{2} x+2}\ \right|+C$$
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Subjective Medium Published on 17th 09, 2020
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