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Indefinite Integration — Maths STD 12 Science — Question

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

Consider, $\frac{2 x^2-3}{\left(x^2-5\right)\left(x^2+4\right)}$
Let
$
\begin{array}{ll}
\text { Let } & x^2=m \\
\therefore & =\frac{2 m-3}{(m-5)(m+4)} \ldots \text { proper rational function. }
\end{array}
$
Now, $\frac{2 m-3}{(m-5)(m+4)}=\frac{A}{(m-5)}+\frac{B}{(m+4)}=\frac{A(m+4)+B(m-5)}{(m-5)(m+4)}$
$
\begin{aligned}
& \therefore \quad 2 m-3=A(m+4)+B(m-5) \\
& \text { at } m=5, \quad 2(5)-3 \quad=A(9)+B(0) \\
& 7=9 A \quad \Rightarrow A=\frac{7}{9} \\
& \text { at } m=-4, \quad 2(-4)-3=A(0)+B(-9) \\
& -11=-9 B \Rightarrow B=\frac{11}{9}
\end{aligned}
$
Thus, $\frac{2 m-3}{(m-5)(m+4)}=\frac{\left(\frac{7}{9}\right)}{(m-5)}+\frac{\left(\frac{11}{9}\right)}{(m+4)} \quad$ i.e. $\frac{2 x^2-3}{\left(x^2-5\right)\left(x^2+4\right)}=\frac{\left(\frac{7}{9}\right)}{x^2-5}+\frac{\left(\frac{11}{9}\right)}{x^2+4}$
$
\begin{aligned}
& \therefore \mathrm{I}=\int\left[\frac{\left(\frac{7}{9}\right)}{x^2-5}+\frac{\left(\frac{11}{9}\right)}{x^2+4}\right] \cdot d x \\
& =\frac{7}{9} \cdot \int \frac{1}{x^2-(\sqrt{5})^2} \cdot d x+\frac{11}{9} \cdot \int \frac{1}{x^2+(2)^2} \cdot d x \\
& =\frac{7}{9} \cdot \frac{1}{2(\sqrt{5})} \cdot \log \left[\frac{x-\sqrt{5}}{x+\sqrt{5}}\right]+\frac{11}{9} \cdot \frac{1}{2} \cdot \tan ^{-1}\left(\frac{x}{2}\right)+c \\
& \therefore \mathrm{I}=\frac{7}{18(\sqrt{5})} \cdot \log \left[\frac{x-\sqrt{5}}{x+\sqrt{5}}\right]+\frac{11}{18} \cdot \tan ^{-1}\left(\frac{x}{2}\right)+c \\
& \therefore \int \frac{2 x^2-3}{\left(x^2-5\right)\left(x^2+4\right)} \cdot d x=\frac{7}{18(\sqrt{5})} \cdot \log \left[\frac{x-\sqrt{5}}{x+\sqrt{5}}\right]+\frac{11}{18} \cdot \tan ^{-1}\left(\frac{x}{2}\right)+c \\
&
\end{aligned}
$

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