Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals5 Marks
Question
Integrate the function $\frac{x+3}{x^{2}-2 x-5}$
✓
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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