Mathematics

Integrate the following functions w.r.t $$X :\dfrac{7+4x+5x^2}{(2x+3)^{\dfrac{3}{2}}}$$


SOLUTION
Let $$I=\displaystyle \int \dfrac{7+4x+5x^2}{(2x+3)^{\dfrac{3}{2}}}.dx$$

$$=\dfrac{5x^2+4x+7}{(2x+3)^{\dfrac{3}{2}}}.dx$$

Put $$2x+3=t$$
$$\therefore 2dx=dt$$
$$\therefore dx=\dfrac{dt}{2}$$
Also, $$x=\dfrac{t-3}{2}$$
$$ \therefore I=\displaystyle \int \dfrac{5 \left(\dfrac{t-2}{2}\right)^2+4 \left(\dfrac{t-3}{2}\right)+7}{t^{\dfrac{3}{2}}}.\dfrac{dt}{2}$$

$$=\dfrac{1}{2} \displaystyle \int$$$$ \dfrac{5 \left(\dfrac{t^2-6t+9}{4}\right)+2(t-3)+7}{t^{\dfrac{3}{2}}} dt$$

$$=\dfrac{1}{2} \displaystyle \int \dfrac{5 t^2-30 t+45+8t-24+28}{4t^{\dfrac{3}{2}}} dt$$
$$=\dfrac{1}{8} \displaystyle \int \dfrac{5t^2-22t+49}{t^{\dfrac{3}{2}}} dt$$
$$=\dfrac{1}{8} \displaystyle \int$$$$ \left(5t^{\dfrac{1}{2}}-22t^{-\dfrac{1}{2}}+49t^{-\dfrac{3}{2}} \right) dt$$
$$\dfrac{5}{8} \displaystyle \int t^{\dfrac{1}{2}}dt-\dfrac{22}{8} \displaystyle \int t^{-\dfrac{1}{2}}dt+\dfrac{49}{8} \displaystyle \int t^{-\dfrac{3}{2}}dt$$
$$=\dfrac{5}{8}. \dfrac{t^{-\dfrac{3}{2}}}{\left(t^{-\dfrac{3}{2}} \right)}-\dfrac{11}{4}.\dfrac{t^{-\dfrac{1}{2}}}{\left(t^{-\dfrac{1}{2} }\right)}+\dfrac{49}{8}. \dfrac{t^{-\dfrac{1}{2}}}{\left( t^{-\dfrac{3}{2}}\right)}+c$$
$$\dfrac{5}{12}(x+3)^{\dfrac{3}{2}}-\dfrac{11}{2} \sqrt{2x+3}-\dfrac{49}{4}.\dfrac{1}{\sqrt{2x+3}}+c$$
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Subjective Medium Published on 17th 09, 2020
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