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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

#### Realted Questions

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$\int (sin^{-1} x + cos^{-1} x) dx =$
• A. $x (sin6{-1} x - cos^{-1} x) + c$
• B. $x (cos^{-1} x - sin^{-1} x ) + c$
• C. $\dfrac{\pi}{2} + x + c$
• D. $\dfrac{1}{2} \pi x + c$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve:
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Q3 Subjective Medium
Calculate the following integral:
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Q4 Subjective Hard
Calculate the following integrals.
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