Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 33 Marks
Question
Evaluate: $\int \frac{x+2}{\sqrt{x^2+2 x-1}} d x$
✓
Answer
Let the given integral be,
$l=\int \frac{x+2}{\sqrt{x^2+2 x-1}} d x$
Let $x +2=\lambda \frac{d}{d x}\left( x ^2+2 x -1\right)+\mu$
$ x +2=\lambda(2 x + x )+\mu$
$x +2=(2 \lambda) x +2 \lambda+\mu$
Comparing the coefficients of like powers of $x,$
$2 \lambda=1 $
$\Rightarrow \lambda=\frac{1}{2}$
$2 \lambda+\mu=2$
$\Rightarrow 2\left(\frac{1}{2}\right)+\mu=2$
$\mu=1$
So $, I _1=\int \frac{\frac{1}{2}(2 x+2)+1}{\sqrt{x^2+2 x-1}} d x$
$=\frac{1}{2} \int \frac{1}{\sqrt{x^2+2 x-1}}(2 x +2) dx +1 \frac{1}{\sqrt{x^2+2 x+(1)^2-(1)^2-1}} d x$
$I =\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^2+2 x-1}} d x+1 \frac{1}{(x+1)^2-(\sqrt{2})^2} d x$
$I =\frac{1}{2} 2 \sqrt{x^2+2 x-1}+\log \left| x +1+\sqrt{(x+1)^2-(\sqrt{2})^2}\right|+c\left]\right.$
$I =\sqrt{x^2+2 x-1}+\log \left| x +1+\sqrt{x^2+2 x-1}\right|+ c$
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