Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 83 Marks
Question
Evaluate: $\int \frac{(2 x+3)}{\sqrt{x^2+x+1}} d x$
✓
Answer
Formula to be used $-\int \frac{d x}{\sqrt{x^2 \pm a^2}}=\log \left(x+\sqrt{x^2 \pm a^2}\right)+c$ where $c$ is the integrating constant
$\therefore \int \frac{(2 x+3)}{\sqrt{x^2+x+1}} d x$
$=\int \frac{(2 x+1)+2}{\sqrt{x^2+x+1}} d x$
$=\int \frac{(2 x+1)}{\sqrt{x^2+x+1}} d x+\int \frac{2}{\sqrt{x^2+x+1}} d x$
Put$, x ^2+ x +1= a ^2,(2 x +1) dx =2 ada$
$\therefore \int \frac{(2 x+1)}{\sqrt{x^2+x+1}} dx$
$=\int \frac{2 d a}{a}$
$=\int 2 da$
$=2 a + c _1$
$=2 \sqrt{x^2+x+1}+c_1$
For $2^{nd}$ part of integral.
$\therefore \int \frac{2}{\sqrt{x^2+x+1}} d x$
$=2 \int \frac{d x}{\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}$
$=2 \log \left|\left(x+\frac{1}{2}\right)+\sqrt{ x ^2+ x +1}\right|+ c _2$
$\therefore \int \frac{(2 x +1)}{\sqrt{ x ^2+ x +1}} dx +\int \frac{2}{\sqrt{x^2+x+1}} d x$
$=2 \sqrt{x^2+x+1}+2 \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x+1}\right|+c, c$ is the integrating constant
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