Question Bank [2022] — Maths STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceMathsQuestion Bank [2022]4 Marks
Question
$\int \frac{(2 \log x+3)}{x(3 \log x+2)\left[(\log x)^2+1\right]} d x$
✓
Answer
$\text { Let } I =\int \frac{(2 \log x+3)}{x(3 \log x+2)\left[(\log x)^2+1\right]} d x$
Put $\log x=t$
$ \therefore \frac{1}{x} d x=d t$
$\therefore I =\int \frac{2 t +3}{(3 t +2)\left( t ^2+1\right)} dt $
Let $\frac{2+3}{(3 t+2)\left(t^2+1\right)}=\frac{A}{3 t+2}+\frac{B t+C}{t^2+1}$
$\therefore 2 t+3=A\left(t^2+1\right)+(B t+C)(3 t+2)\ldots(i)$
Putting $t =-\frac{2}{3}$ in $( i )$, we get
$ 2\left(\frac{-2}{3}\right)+3= A \left[\left(\frac{-2}{3}\right)^2+1\right]$
$\therefore \frac{-4}{3}+3= A \left(\frac{4}{9}+1\right)$
$\therefore \frac{5}{3}= A \left(\frac{13}{9}\right)$
$\therefore A =\frac{15}{13} $
Putting $t=0$ in (i), we get
$ 3= A (1)+ C (2)$
$\therefore 3=\frac{15}{13}+2 C$
$\therefore 3-\frac{15}{13}=2 C$
$\therefore \frac{24}{13}=2 C$
$\therefore C =\frac{12}{13} $
Putting $t=1$ in (i), we get
$ 2+3=A(1+1)+(B+C)(3+2)$
$\therefore 5=2 A+5(B+C)$
$\therefore 5=2\left(\frac{15}{13}\right)+5\left(B+\frac{12}{13}\right)$
$\therefore 5=\frac{30}{13}+5 B+\frac{60}{13}$
$\therefore 5 B=5-\frac{30}{13}-\frac{60}{13}$
$\therefore 5 B=-\frac{25}{13}$
$\therefore B=\frac{-5}{13} $
$\therefore \frac{2 t +3}{(3 t +2)\left( t ^2+1\right)}=\frac{\frac{15}{13}}{3 t +2}+\frac{-\frac{5}{13} t +\frac{12}{13}}{ t ^2+1}$
$\therefore I =\int\left(\frac{\frac{15}{13}}{3 t +2}+\frac{\frac{-5}{13} t +\frac{12}{13}}{ t ^2+1}\right) dt$
$=\frac{15}{13} \int \frac{1}{3 t +2} dt -\frac{5}{13} \int \frac{ t }{ t ^2+1} dt +\frac{12}{13} \int \frac{1}{ t ^2+1} dt$
$=\frac{15}{13} \int \frac{1}{3 t +2} dt -\frac{5}{13} \cdot \frac{1}{2} \int \frac{2 t }{ t ^2+1} dt +\frac{12}{13} \int \frac{1}{ t ^2+1} dt$
$=\frac{15}{13} \cdot \frac{\log |3 t +2|}{3}-\frac{5}{26} \log \left| t ^2+1\right|+\frac{12}{13} tan ^{-1} t + c$
$\therefore I =\frac{5}{13} \log |3 \log x 2|-\frac{5}{26} \log \left|(\log x)^2+1\right|+\frac{12}{13} \tan ^{-1}(\log x)+ c $
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