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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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