$\int \frac{2 x^2-1}{\left(x^2+4\right)\left(x^2-3\right)} d x=$

$\int \frac{2 x^2-1}{\left(x^2+4\right)\left(x^2-3\right)} d x=$
  1. $\frac{9}{14} \tan ^{-1}\left(\frac{x}{2}\right)+\frac{5}{14 \sqrt{3}} \log \left(\frac{x-\sqrt{3}}{x+\sqrt{3}}\right)+\mathrm{c}$, (where c is constant of integration)
  2. $\frac{9}{7} \tan ^{-1}\left(\frac{x}{2}\right)+\frac{5}{7 \sqrt{3}} \log \left(\frac{x-\sqrt{3}}{x+\sqrt{3}}\right)+\mathrm{c}$, (where c is constant of integration)
  3. $\frac{9}{7} \tan ^{-1}\left(\frac{x}{2}\right)-\frac{5}{7 \sqrt{3}} \log \left(\frac{x-\sqrt{3}}{x+\sqrt{3}}\right)+\mathrm{c}$, (where c is constant of integration)
  4. $\frac{9}{14} \tan ^{-1}\left(\frac{x}{2}\right)+\frac{5}{7} \log \left(\frac{x-\sqrt{3}}{x+\sqrt{3}}\right)+\mathrm{c}$, (where c is constant of integration)

Solution

$\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{2 x^2-1}{\left(x^2+4\right)\left(x^2-3\right)} \mathrm{d} x \\ & \frac{2 x^2-1}{\left(x^2+4\right)\left(x^2-3\right)}=\frac{\mathrm{A}}{x^2+4}+\frac{\mathrm{B}}{x^2-3} \\ \therefore \quad & 2 x^2-1=\mathrm{A}\left(x^2-3\right)+\mathrm{B}\left(x^2+4\right) \\ \therefore \quad & 2 x^2-1=(\mathrm{A}+\mathrm{B}) x^2-(3 \mathrm{~A}-4 \mathrm{~B})\end{aligned}$ $\begin{aligned} & \Rightarrow A+B=2 \text { and } 3 A-4 B=1 \\ & \Rightarrow A=\frac{9}{7} \text { and } B=\frac{5}{7}\end{aligned}$ $\therefore \quad \mathrm{I}=\frac{9}{7} \int \frac{1}{x^2+(2)^2} \mathrm{~d} x+\frac{5}{7} \int \frac{1}{x^2-(\sqrt{3})^2} \mathrm{~d} x$ $\begin{aligned} & =\frac{9}{7} \times \frac{1}{2} \tan ^{-1}\left(\frac{x}{2}\right)+\frac{5}{7} \times \frac{1}{2 \sqrt{3}} \log \left(\frac{x-\sqrt{3}}{x+\sqrt{3}}\right)+\mathrm{c} \\ & =\frac{9}{14} \tan ^{-1}\left(\frac{x}{2}\right)+\frac{5}{14 \sqrt{3}} \log \left(\frac{x-\sqrt{3}}{x+\sqrt{3}}\right)+\mathrm{c}\end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 1)

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