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Integration (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Integration (p-1)5 Marks
Question
Evalute : $\int \frac{2 x^3-3 x^2-9 x+1}{2 x^2-x-10} d x$
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Answer

$
\begin{aligned}
& \text { Let } I=\int \frac{2 x^3-3 x^2-9 x+1}{2 x^2-x-10} d x \\
&\left.2 x^2-x-10\right) 2 x^3-3 x^2-9 x+1(x-1 \\
& \frac{2 x^3-x^2-10 x}{+}+ \\
&-2 x^2+x+1 \\
&+2 x^2+x+10 \\
&+-9
\end{aligned}
$
$
\begin{array}{rl}
\therefore I & 2 x^3-3 x^2-9 x+1=(x-1)\left(2 x^2-x-10\right)-9 \\
\therefore I & =\int\left[\frac{(x-1)\left(2 x^2-x-10\right)-9}{2 x^2-x-10}\right] d x \\
& =\int\left[(x-1)-\frac{9}{2 x^2-x-10}\right] d x \\
& =\int(x-1) d x-\frac{9}{2} \int \frac{1}{x^2-\frac{1}{2} x-5} d x \\
& =\int x d x-\int 1 d x-\frac{9}{2} \int \frac{1}{\left(x^2-\frac{1}{2} x+\frac{1}{16}\right)^{-}-\frac{1}{16}-5} d x \\
& =\int x d x-\int 1 d x-\frac{9}{2} \int \frac{1}{\left(x-\frac{1}{4}\right)^2-\left(\frac{9}{4}\right)^2} d x
\end{array}
$
$
=\frac{x^2}{2}-x-\frac{9}{2} \times \frac{1}{2 \times \frac{9}{4}} \log \left|\frac{x-\frac{1}{4}-\frac{9}{4}}{x-\frac{1}{4}+\frac{9}{4}}\right|+c_1
$
$\begin{aligned} & =\frac{x^2}{2}-x-\log \left|\frac{x-\frac{5}{2}}{x+2}\right|+c_1 \\ & =\frac{x^2}{2}-x-\log \left|\frac{2 x-5}{2(x+2)}\right|+c_1 \\ & =\frac{x^2}{2}-x+\log \left|\frac{2(x+2)}{2 x-5}\right|+c_1 \\ & =\frac{x^2}{2}-x+\log \left|\frac{x+2}{2 x-5}\right|+\log 2+c_1 \\ & =\frac{x^2}{2}-x+\log \left|\frac{x+2}{2 x-5}\right|+c, \text { where } c_1=\log 2+c_1\end{aligned}$

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