Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 83 Marks
Question
Evaluate the integral: $\int(x+1) \sqrt{x^2+x+1} d x$
✓
Answer
Let the given integral be,
$ I =\int(x+1) \sqrt{x^2+x+1} d x$
Also$, x +1=\lambda \frac{d}{d x}\left(x^2+x+1\right)+\mu$
$\Rightarrow x +1=\lambda(2 x +1)+\mu$
$\Rightarrow x +1=(2 \lambda) x +\lambda+\mu$
Eqyating coeffient of like terms
$2 \lambda=1$
$\Rightarrow \lambda=\frac{1}{2}$
And
$\lambda+\mu=1$
$\Rightarrow \frac{1}{2}+\mu=1$
$\therefore \mu=\frac{1}{2}$
$\therefore I =\frac{1}{2} \int(2 x+1) \sqrt{x^2+x+1} d x+\frac{1}{2} \int \sqrt{x^2+x+1} d x$
$=\frac{1}{2} \int(2 x+1) \sqrt{x^2+x+1} d x+\frac{1}{2} \int \sqrt{x^2+x+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1} d x$
$=\frac{1}{2} \int(2 x+1) \sqrt{x^2+x+1} d x+\frac{1}{2} \int \sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} d x$
Let $x ^2+ x +1= t$
$\Rightarrow(2 x +1) dx = dt$
Then,
$I=\frac{1}{2} \int \sqrt{t} d t+\frac{1}{2}\left[\frac{x+\frac{1}{2}}{2} \sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}+\frac{3}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right|\right]+C$
$=\frac{1}{2} \times \frac{2}{3} t^{\frac{3}{2}}+\frac{1}{2}\left[\left(\frac{2 x+1}{4}\right) \sqrt{x^2+x+1}+\frac{3}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x+1}\right|\right]+C$
$=\frac{1}{3}\left(x^2+x+1\right)^{\frac{3}{2}}+\frac{1}{2}\left[\left(\frac{2 x+1}{4}\right) \sqrt{x^2+x+1}+\frac{3}{8} \log \left|\left(x+\frac{1}{2}\right)+\sqrt{x^2+x+1}\right|\right]+C$
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