Solve the Following Question.(4 Marks)

Question 14 Marks

Evalute : $\int \frac{5 x^2-6 x+3}{2 x-3} d x$

Answer

Let $I=\int \frac{5 x^2-6 x+3}{2 x-3} d x$
$
\begin{aligned}
2 x - 3 \longdiv { 5 x ^ { 2 } - 6 x + 3 } \frac { 5 } { 2 } x + \frac { 3 } { 4 } \\
5 x^2-\frac{15}{2} x \\
\frac{-\quad+}{\frac{3}{2} x+3} \\
\frac{3}{2} x-\frac{9}{4} \\
\frac{-\quad+}{\frac{21}{4}} \\
\therefore 5 x^2-6 x+3=\left(\frac{5}{2} x+\frac{3}{4}\right)(2 x-3)+\frac{21}{4} \\
\therefore I=\int\left[\frac{\left(\frac{5}{2} x+\frac{3}{4}\right)(2 x-3)+\frac{21}{4}}{2 x-3}\right] d x \\
=\int\left[\frac{5}{2} x+\frac{3}{4}+\frac{\left(\frac{21}{4}\right)}{2 x-3}\right] d x \\
=\frac{5}{2} \int x d x+\frac{3}{4} \int 1 d x+\frac{21}{4} \int \frac{1}{2 x-3} d x \\
=\frac{5}{2} \cdot \frac{x^2}{2}+\frac{3}{4} x+\frac{21}{4} \cdot \frac{\log |2 x-3|}{2}+c \\
=\frac{5 x^2}{4}+\frac{3 x}{4}+\frac{21}{8} \\
=\frac{5 x^2}{4}+\frac{3 x}{4}+\frac{21}{8} \log |2 x-3|+c \text {. } \\
\end{aligned}
$

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Question 24 Marks

Evalute : $\int \frac{x^2+x-1}{x^2+x-6} d x$

Answer

$
\text { Let } \begin{aligned}
I & =\int \frac{x^2+x-1}{x^2+x-6} d x \\
& =\int \frac{\left(x^2+x-6\right)+5}{x^2+x-6} d x \\
& =\int\left[1+\frac{5}{x^2+x-6}\right] d x \\
& =\int 1 d x+5 \int \frac{1}{x^2+x-6} d x
\end{aligned}
$
Let $\frac{1}{x^2+x-6}=\frac{1}{(x+3)(x-2)}=\frac{A}{x+3}+\frac{B}{x-2}$
$
\therefore 1= A ( x -2)+ B ( x +3)
$
Put $x+3=0$, i.e. $x=-3$, we get
$
\begin{aligned}
& 1= A (-5)+ B (0) \\
& \therefore A =\frac{-1}{5}
\end{aligned}
$
Put $x-2=0$, i.e. $x=2$, we get
$
\begin{aligned}
& 1= A (0)+ B (5) \\
& \therefore B =\frac{1}{5} \\
& \therefore \frac{1}{x^2+x-6}=\frac{(-1 / 5)}{x+3}+\frac{(1 / 5)}{x-2} \\
& \therefore I=\int 1 d x+5 \int\left[\frac{(-1 / 5)}{x+3}+\frac{(1 / 5)}{x-2}\right] d x \\
& \quad=\int 1 d x-\int \frac{1}{x+3} d x+\int \frac{1}{x-2} d x
\end{aligned}
$

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Maths (commerce) Solve the Following Question.(4 Marks)

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