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

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Integration (p-1)4 Marks
Question
Evalute : $\int \frac{5 x^2-6 x+3}{2 x-3} d x$
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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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