Let I = 1 x(3+2 x) d x = x d x ^2 x(3+2 x) = x d x (1- ^2 x )(3+2 x) = x d x (1+ x)(1- x)(3+2 x) Put x=t - xd x= dt I = -1 (1+ t )(1- t )(3+2 t ) dt Let 1 (1+ t )(1- t )(3+2 t ) =A 1+t +B 1-t +C 3+2 t -1=A(1-t)(3+2 t)+B(1+t)(3+2 t)+C(1+t)(1-t) (i) Putting t=1 in (i), we get -1=10 B B=-1 10 Putting t=-1 in (i), we get -1=2 A A=-1 2 Putting t=-3 2 in (i), we get -1=-5 / 4 "C" C=4 5 -1 (1+ t )(1- t )(3+2 t ) = -1 2 1+ t + -1 10 1- t + -4 5 3+2 t I = [-1 2(1+ t ) +(-1) 10(1- t ) +4 5(3+2 t ) ] dt =-1 2 1 1+ t dt -1 10 1 1- t dt +4 5 1 3+2 t dt =-1 2 |1+ t |-1 10 |1- t | -1 +4 5 |3+2 t | 2 + c I =-1 2 |1+ x|+1 10 |1- x|+2 5 |3+2 x|+ c

1 x(3+2 x) d x

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$\int \frac{1}{\sin x(3+2 \cos x)} d x$

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Answer

$ \text { Let } I =\int \frac{1}{\sin x(3+2 \cos x)} d x$
$=\int \frac{\sin x d x}{\sin ^2 x(3+2 \cos x)}$
$=\int \frac{\sin x d x}{\left(1-\cos ^2 x\right)(3+2 \cos x)}$
$=\int \frac{\sin x d x}{(1+\cos x)(1-\cos x)(3+2 \cos x)} $
Put $\cos x=t$
$ \therefore-\sin xd x= dt$
$\therefore I =\int \frac{-1}{(1+ t )(1- t )(3+2 t )} dt $
Let $\frac{1}{(1+ t )(1- t )(3+2 t )}$
$ =\frac{A}{1+t}+\frac{B}{1-t}+\frac{C}{3+2 t}$
$\therefore-1=A(1-t)(3+2 t)+B(1+t)(3+2 t)+C(1+t)(1-t)\ldots(i) $
Putting $t=1$ in (i), we get
$ -1=10 B$
$\therefore B=\frac{-1}{10} $
Putting $t=-1$ in (i), we get
$ -1=2 A$
$\therefore A=\frac{-1}{2} $
Putting $t=-\frac{3}{2}$ in (i), we get
$ -1=-5 / 4 \text { "C" }$
$\therefore C=\frac{4}{5}$
$\therefore \frac{-1}{(1+ t )(1- t )(3+2 t )}=\frac{\frac{-1}{2}}{1+ t }+\frac{\frac{-1}{10}}{1- t }+\frac{\frac{-4}{5}}{3+2 t }$
$\therefore I =\int\left[\frac{-1}{2(1+ t )}+\frac{(-1)}{10(1- t )}+\frac{4}{5(3+2 t )}\right] dt$
$=-\frac{1}{2} \int \frac{1}{1+ t } dt -\frac{1}{10} \int \frac{1}{1- t } \cdot dt +\frac{4}{5} \int \frac{1}{3+2 t } dt$
$=\frac{-1}{2} \log |1+ t |-\frac{1}{10} \cdot \frac{\log |1- t |}{-1}+\frac{4}{5} \cdot \frac{\log |3+2 t |}{2}+ c$
$\therefore I =\frac{-1}{2} \log |1+\cos x|+\frac{1}{10} \log |1-\cos x|+\frac{2}{5} \log |3+2 \cos x|+ c $

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