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Question Bank [2022] — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]3 Marks
Question
$\int \frac{7+4 x+5 x^2}{(2 x+3)^{\frac{3}{2}}} d x$
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Answer

$Lt I =\int \frac{x^2+4 x+7}{(2 x+3)^{\frac{3}{2}}} d x$
Put $2 x+3=t^2$$\ldots(i)$
Differentiating w.r.t. $x$, we get
$ 2 dx =2 t d t$
$\therefore dx = t d t $
From (i), we get
$ x =\frac{ t ^2-3}{2}$
$\therefore I =\int \frac{5\left(\frac{ t ^2-3}{2}\right)^2+4\left(\frac{ t ^2-3}{2}\right)+7}{\left( t ^2\right)^{\frac{3}{2}}} \cdot t d t$
$=\int \frac{5\left(\frac{ t ^4-6 t ^2+9}{4}\right)+2 t ^2-6+7}{ t ^3} \cdot t d t$
$=\int \frac{5 t ^4-30 t ^2+45+8 t ^2+4}{4 t ^3} \cdot t d t$
$=\int \frac{5 t ^4-22 t ^2+49}{4 t ^2} dt$
$=\frac{5}{4} \int t ^2 dt -\frac{22}{4} \int dt +494 \int t ^{-2} dt$
$=\frac{5}{4} \cdot \frac{ t ^3}{3}-\frac{22}{4} t +\frac{49}{4} \cdot \frac{ t ^{-1}}{-1}+ c$
$=\frac{5}{12} t ^3-\frac{11}{2} t -\frac{49}{4 t }+ c$
$\therefore I =\frac{5}{12}(2+3)^{\frac{3}{2}}-\frac{11}{2} \sqrt{2 x+3}-\frac{49}{4} \cdot \frac{1}{\sqrt{2 x+3}}+ c$

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