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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals5 Marks
Question
Integrate the function $\frac{x+3}{x^{2}-2 x-5}$
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Answer

Let $x + 3 = A \frac{d}{d x}\left(x^{2}-2 x-5\right)+B$
$\Rightarrow x + 3 = A(2x - 2) + B$
Now, equating the coefficients of $x$ and constant term on both sides, we get,
$2A = 1$
$\Rightarrow A=\frac{1}{2}$
$-2A + B = 3$
$\Rightarrow B = 4$
$\Rightarrow x+3=\frac{1}{2}(2 x-2)+4$
Now, $\int \frac{x+3}{x^{2}-2 x-5} d x=\int \frac{\frac{1}{2}(2 x-2)+4}{x^{2}-2 x-5} d x$
$=\frac{1}{2} \int \frac{2 x-2}{x^{2}-2 x-5} d x+4 \int \frac{1}{x^{2}-2 x-5} d x$
Now, Let us consider $\int \frac{2 x-2}{x^{2}-2 x-5} d x$
Let $x^2 - 2x - 5 = t$
$\Rightarrow (2x - 2)dx = dt$
$\therefore \int \frac{2 x-2}{x^{2}-2 x-5} d x=\int \frac{d t}{t}=\log |t|=\log \left|x^{2}-2 x-5\right| .......(i)$
And, now let us consider, $\int \frac{1}{x^{2}-2 x-5} d x$
$\Rightarrow \int \frac{1}{\left(x^{2}-2 x+1\right)-6} d x$
$\Rightarrow \int \frac{1}{(x-1)^{2}+(\sqrt{6})^{2}} d x$
$=\frac{1}{2 \sqrt{6}} \log \left(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right)......(ii)$
Using eq. $(i)$ and $(ii),$ we get,
$\Rightarrow \int \frac{x+3}{x^{2}-2 x-5} d x=\frac{1}{2} \log \left|x^{2}-2 x-5\right|+\frac{4}{2 \sqrt{6}} \log \left(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right)+C$ 
$\Rightarrow \int \frac{x+3}{x^{2}-2 x-5} d x=\frac{1}{2} \log \left|x^{2}-2 x-5\right|+\frac{2}{\sqrt{6}} \log \left(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right)+C$

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