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Indefinite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integration5 Marks
Question
Evaluate : $\int \frac{2 x-3}{3 x^2+4 x+5} \cdot d x$
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Answer

$\quad 2 x-3=A \cdot \frac{d}{d x}\left(3 x^2+4 x+5\right)+B$
$
\begin{aligned}
2 x-3 & =A(6 x+4)+B \\
& =(6 A) x+(4 A+B)
\end{aligned}
$
compairing the sides/ the co-efficients of like variables and constants
$
\begin{aligned}
& 6 A=2 \text { and } 4 A+B=-3 \\
& \Rightarrow A=\frac{1}{3} \text { and } B=-\frac{13}{3} \\
& =\int \frac{\frac{1}{3} \cdot \frac{d}{d x}\left(3 x^2+4 x+5\right)+\left(-\frac{13}{3}\right)}{3 x^2+4 x+5} \cdot d x \\
& =\frac{1}{3} \cdot \int \frac{\frac{d}{d x}\left(3 x^2+4 x+5\right)}{3 x^2+4 x+5} \cdot d x-\frac{13}{3} \int \frac{1}{3 x^2+4 x+5} \cdot d x \\
& =\frac{1}{3} \cdot \int \frac{6 x+4}{3 x^2+4 x+5} \cdot d x-\frac{13}{3} \int \frac{1}{3 x^2+4 x+5} \cdot d x \\
& =\mathrm{I}_1-\mathrm{I}_2 \\
& \therefore \quad \mathrm{I}_1=\frac{1}{3} \cdot \int \frac{6 x+4}{3 x^2+4 x+5} \cdot d x \\
& \text { put } 3 x^2+4 x+5=t \\
& \therefore \quad(6 x+4) \cdot d x=1 \cdot d t \\
& \mathrm{I}_1=\frac{1}{3} \cdot \int \frac{1}{t} \cdot d t \\
& =\frac{1}{3} \cdot \log (t)+c_1 \\
& =\frac{1}{3} \cdot \log \left(3 x^2+4 x+5\right)+c_1 \\
\end{aligned}
$
$
\begin{aligned}
& \therefore \quad I_2=\frac{13}{3} \cdot \int \frac{1}{3 x^2+4 x+5} \cdot d x \\
& =\frac{13}{3} \cdot \frac{1}{3} \cdot \int \frac{1}{x^2+\frac{4}{3} x+\frac{5}{3}} \cdot d x \\
& \because\left\{\left(\frac{1}{2} \text { coefficient of } t\right)^2\right. \\
& \left.=\left(\frac{1}{2}\left(\frac{4}{3}\right)\right)^2=\left(\frac{2}{3}\right)^2=\frac{4}{9}\right\} \\
& \mathrm{I}_2=\frac{13}{9} \cdot \int \frac{1}{x^2+\frac{4}{3} x+\frac{4}{9}-\frac{4}{9}+\frac{5}{3}} \cdot d x \\
& =\frac{13}{9} \cdot \int \frac{1}{x^2+\frac{4}{3} x+\frac{4}{9}+\frac{11}{9}} \cdot d x \\
& =\frac{13}{9} \cdot \int \frac{1}{\left(x+\frac{2}{3}\right)^2+\left(\frac{\sqrt{11}}{3}\right)^2} \cdot d x \\
& \because \quad \int \frac{1}{X^2+A^2} \cdot d x=\frac{1}{A} \tan ^{-1}\left(\frac{X}{A}\right)+c \\
& =\frac{13}{9} \cdot \frac{1}{\frac{\sqrt{11}}{3}} \cdot \tan ^{-1}\left(\frac{x+\frac{2}{3}}{\frac{\sqrt{11}}{3}}\right)+c_1 \\
& I_2=\frac{13}{3 \sqrt{11}} \cdot \tan ^{-1}\left(\frac{3 x+2}{\sqrt{11}}\right)+c_2 \ldots \ldots \\
&
\end{aligned}
$
thus, from (i), (ii) and (iii)
$
\begin{aligned}
& \therefore \quad \int \frac{2 x-3}{3 x^2+4 x+5} \cdot d x \\
& =\frac{1}{3} \cdot \log \left(3 x^2+4 x+5\right)-\frac{13}{3 \sqrt{11}} \cdot \tan ^{-1}\left(\frac{3 x+2}{\sqrt{11}}\right)+c \\
& \quad\left(\because c_1+c_2=c\right)
\end{aligned}
$

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