Question Bank [2022] — Maths STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 Marks
Question
$\int \frac{3 x+4}{\sqrt{2 x^2+2 x+1}} d x$
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Answer
$\text { Let } I =\int \frac{3 x+4}{\sqrt{2 x^2+2 x+1}} d x$
Let $3 x +4= A \frac{ d }{ d x}\left(2 x^2+2 x+1\right)+ B$
$ \therefore 3 x+4=A(4 x+2)+B$
$\therefore 3 x+4=4 A x+2 A+B $
By equating the coefficients on both sides, we get
$4 A=3 \text { and } 2 A+B=4$
$\therefore A=\frac{3}{4} \text { and } 2\left(\frac{3}{4}\right)+B=4$
$\therefore B =\frac{5}{2}$
$ \therefore 3 x +4=\frac{3}{4}(4 x+2)+\frac{5}{2}$
$\therefore I =\int \frac{\frac{3}{4}(4 x+2)+\frac{5}{2}}{\sqrt{2 x^2+2 x+1}} d x$
$=\frac{3}{4} \int \frac{4 x+2}{\sqrt{2 x^2+2 x+1}} d x+\frac{5}{2} \int \frac{1}{\sqrt{2 x^2+2 x+1}} d x $ $ =I_1+I_2\ldots(i)$
$I_1=\frac{3}{4} \int \frac{4 x+2}{\sqrt{2 x^2+2 x+1}} d x $
Put $2 x^2+2 x+1=t$
$ \therefore(4 x +2) dx = dt$
$\therefore I _1=\frac{3}{4} \int \frac{ dt }{\sqrt{ t }}$
$=\frac{3}{4} \int t ^{\frac{1}{2}} dt$
$=\frac{3}{4}\left(\frac{ t ^{\frac{1}{2}}}{\frac{1}{2}}\right)+ c _1$
$=\frac{3}{2} \sqrt{ t }+ c _1$
$\therefore I _1=\frac{3}{2} \sqrt{2 x^2+2 x+1}+ c _1\ldots(i)$
$I _2=\frac{5}{2} \int \frac{1}{\sqrt{2 x^2+2 x+1}} d x$
$=\frac{5}{2} \int \frac{1}{\sqrt{2\left(x^2+x+\frac{1}{2}\right)}} d x $
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