Mathematics

$$\int\dfrac{5x^2+4x+7}{(2x+3)^{\frac{3}{2}}}$$


SOLUTION
Let $$I=\int \dfrac{5x^2+4x+7}{(2x+3)^{\frac{3}{2}}}dx$$
Put $$2x+3=t^2$$
Differentiating w.r.t $$x,$$ we get
$$2dx=2t\,dt$$
$$\therefore$$  $$dx=tdt$$
From ( 1 ), we get
$$x=\dfrac{t^2-3}{2}$$
$$\therefore$$  $$I=\int \dfrac{5\left(\dfrac{t^2-3}{2}\right)^2+4\left(\dfrac{t^2-3}{2}\right)+7}{(t^2)^{\frac{3}{2}}}.t\,dt$$

       $$=\int\dfrac{5\left(\dfrac{t^4-6t^2+9}{4}\right)+2t^2-6+7}{t^3}.t\,dt$$

       $$=\int\dfrac{5t^4-30t^2+45+8t^2+4}{4t^3}.t\,dt$$

       $$=\int\dfrac{5t^4-22t^2+49}{4t^2}dt$$

       $$=\dfrac{5}{4}\int t^2dt-\dfrac{22}{4}\int dt+\dfrac{49}{4}\int t^{-2}dt$$

       $$=\dfrac{5}{4}.\dfrac{t^3}{3}-\dfrac{22}{4}t+\dfrac{49}{4}.\dfrac{t^{-2+1}}{-2+1}+c$$

       $$=\dfrac{5}{12}t^3-\dfrac{22}{4}t-\dfrac{49}{4}.t^{-1}+c$$

       $$=\dfrac{5}{12}t^3-\dfrac{22}{4}t-\dfrac{49}{4t}+c$$

$$\therefore$$  $$I=\dfrac{5}{12}(2x+3)^{\frac{3}{2}}-\dfrac{11}{2}\sqrt{2x+3}-\dfrac{49}{4}.\dfrac{1}{\sqrt{2x+3}}+c$$
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Subjective Medium Published on 17th 09, 2020
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