Question Bank [2022] — Maths STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]3 Marks
Question
$\int \frac{6 x^3+5 x^2-7}{3 x^2-2 x-1} d x$
✓
Answer
Let $I =\int \frac{6 x^2+5 x^2-7}{3 x^2-2 x-1} d x$
$\left.3 x^2-2 x-1\right) \frac{2 x +3}{6 x^3+5 x^2+0 x-7}$
$6 x ^3-4 x ^2-2 x$
$\underline{( - )( +) ( +)}$
$9 x ^2+2 x -7$
$9 x ^2-6 x -3$
$\underline{(-)( +)(+)}$
$8 x -4$
$\therefore I =\int\left(2 x+3+\frac{8 x-4}{3 x^2-2 x-1}\right) d x$
$3 x^2-2 x-1=3 x^2-3 x+x-1$
$=3 x(x-1)+1(x-1)$
$=(x-1)(3 x+1)$
$\therefore I =\int\left[2 x+3+\frac{8 x-4}{(x-1)(3 x+1)}\right] d x$
Let $\frac{8 x-4}{(x-1)(3 x+1)}=\frac{ A }{x-1}+\frac{ B }{3 x+1}$
$\therefore 8 x-4=A(3 x+1)+B(x-1)\ldots(i)$
Putting $x=1$ in (i), we get
$4=4 A$
$\therefore A=1$
Putting $x=\frac{-1}{3}$ in (i), we get
$8\left(-\frac{1}{3}\right)-4= B \left(-\frac{1}{3}-1\right)$
$\therefore \frac{-20}{3}=-\frac{4}{3} B$
$\therefore B =5$
$\therefore \frac{8 x-4}{(x-1)(3 x+1)}=\frac{1}{x-1}+\frac{5}{3 x+1}$
$\therefore I =\int\left(2 x+3+\frac{1}{x-1}+\frac{5}{3 x+1}\right) d x$
$=2 \int x d x+3 \int d x+\int \frac{1}{x-1} d x+\frac{5}{3} \int \frac{3}{3 x+1} d x$
$=2\left(\frac{x^2}{2}\right)+3 x+\log |x+1|+\frac{5 \log |3 x+1|}{3}+ c$
$\therefore I =x^2+3 x+\log |x-1|+\frac{5}{3} \log |3 x+1|+ c$
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