Solve the Following Question.(5 Marks)

Question 15 Marks

$\int_1^2 \frac{x+3}{x(x+2)} d x$

Answer

Let $I =\int_1^2 \frac{x+3}{x(x+2)} d x$
Let $\frac{x+3}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}$
$\therefore x +3= A ( x +2)+ Bx$
Put $x=0$, we get
$ 3= A (2)+ B (0) $
$ \therefore A =\frac{3}{2}$\
Put $x+2=0$, i.e. $x=-2$, we get
$ -2+3= A (0)+ B (-2) $
$ \therefore 1=-2 B $
$ \therefore B =-\frac{1}{2}$
$\therefore \frac{x+3}{x(x+2)}=\frac{\left(\frac{3}{2}\right)}{x}+\frac{\left(-\frac{1}{2}\right)}{x+2}$
$\therefore I =\int_1^2\left[\frac{\left(\frac{3}{2}\right)}{x}+\frac{\left(-\frac{1}{2}\right)}{x+2}\right] d x $
$ =\frac{3}{2} \int_1^2 \frac{1}{x} d x-\frac{1}{2} \int_1^2 \frac{1}{x+2} d x $
$ =\frac{3}{2}[\log |x|]_1^2-\frac{1}{2}[\log |x+2|]_1^2$
$ =\frac{3}{2}(\log 2-\log 1)-\frac{1}{2}(\log 4-\log 3)$
$ =\frac{3}{2} \log 2-\frac{1}{2} \log 4+\frac{1}{2} \log 3 \quad \ldots[\because \log 1=0] $
$ \frac{1}{2}(3 \log 2-\log 4+\log 3)$
$=\frac{1}{2}(\log 8-\log 4+\log 3) $
$=\frac{1}{2} \log \left(\frac{8 \times 3}{4}\right)=\frac{1}{2} \log 6 .$

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Question 25 Marks

$\int_1^2 \frac{5 x^2}{x^2+4 x+3} d x$

Answer

Let $I=\int_1^2 \frac{5 x^2}{x^2+4 x+3} d x$
$
=\int_1^2\left[\frac{5\left(x^2+4 x+3\right)-5(4 x+3)}{x^2+4 x+3}\right] d x
$
$
=\int_1^2\left[5-\frac{20 x+15}{x^2+4 x+3}\right] d x
$
$
\therefore I=\int_1^2 5 d x-\int_1^2 \frac{20 x+15}{x^2+4 x+3} d x
$
Let $\frac{20 x+15}{x^2+4 x+3}=\frac{20 x+15}{(x+1)(x+3)}=\frac{A}{x+1}+\frac{B}{x+3}$
$
\therefore 20 x+15=A(x+3)+B(x+1)
$
Put $x+1=0$, i.e. $x=-1$, we get
$
\begin{aligned}
& 20(-1)+15=A(2)+B(0) \\
& \therefore-5=2 A \quad \therefore A=-\frac{5}{2}
\end{aligned}
$
Put $x+3=0$, i.e. $x=-3$, we get
$
20(-3)+15=A(0)+B(-2)
$
$
\therefore-45=-2 B \quad \therefore B=\frac{45}{2}
$
$
\therefore \frac{20 x+15}{x^2+4 x+3}=\frac{\left(-\frac{5}{2}\right)}{x+1}+\frac{\left(\frac{45}{2}\right)}{x+3}
$
$\therefore$ from (1),
$
I=\int_1^2 5 d x-\int_1^2\left[\frac{\left(-\frac{5}{2}\right)}{x+1}+\frac{\left(\frac{45}{2}\right)}{x+3}\right] d x
$
$\begin{aligned} & =5 \int_1^2 1 d x+\frac{5}{2} \int_1^2 \frac{1}{x+1} d x-\frac{45}{2} \int_1^2 \frac{1}{x+3} d x \\ & =5[x]_1^2+\frac{5}{2}[\log |x+1|]_1^2-\frac{45}{2}[\log |x+3|]_1^2 \\ & =5(2-1)+\frac{5}{2}(\log 3-\log 2)-\frac{45}{2}(\log 5-\log 4) \\ & =5+\frac{1}{2}(5 \log 3-5 \log 2-45 \log 5+90 \log 2) \\ & =5+\frac{1}{2}(5 \log 3+85 \log 2-45 \log 5) .\end{aligned}$

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Question 35 Marks

$\int_2^3 \frac{x}{(x+2)(x+3)} d x$

Answer

Let $I =\int_2^3 \frac{x}{(x+2)(x+3)} \cdot d x$
Let $\frac{x}{(x+2) x+3}=\frac{ A }{x+2}+\frac{ B }{x+3}$
$
\therefore x = A ( x +3)+ B ( x +2)
$
Putting $x=-3$ in (ii) we get
$
\begin{aligned}
& -2= A \\
& \therefore B =3
\end{aligned}
$
Putting $x=-2$ in (ii), we get
$
\begin{aligned}
& -2= A \\
& \therefore A =-2
\end{aligned}
$
From (i), we get
$
\begin{aligned}
& \frac{x}{(x+2(x+3))}=\frac{-2}{x+2}+\frac{3}{x+3} \\
& \therefore I =\int_2^3\left[\frac{-2}{x+2}+\frac{3}{x+3}\right] \cdot d x \\
& =-2 \int_2^3 \frac{1}{x+2} \cdot d x+3 \int_2^3 \frac{1}{x+3} \cdot d x \\
& =-2[\log |x+2|]_2^3+3[\log |x+3|]_2^3 \\
& =-2 \log [\log 5-\log 4]+3[\log 6-\log 5] \\
& =-2\left[\log \left(\frac{5}{4}\right)\right]+3\left[\log \left(\frac{6}{5}\right)\right] \\
& =3 \log \left(\frac{6}{5}\right)-2 \log \left(\frac{5}{4}\right)
\end{aligned}
$
$\begin{aligned} & =\log \left(\frac{6}{5}\right)^2-2 \log \left(\frac{5}{4}\right)^2 \\ & =\log \left(\frac{216}{125}\right)-\log \left(\frac{25}{16}\right) \\ & =\log \left(\frac{216}{125} \times \frac{16}{25}\right) \\ & \therefore I =\log \left(\frac{3456}{3125}\right) .\end{aligned}$

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Maths (commerce) Solve the Following Question.(5 Marks)

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