Question

If $\int \frac{\left(x+x^2+x^3\right)}{\sqrt{15+12 x+10 x^2}} d x=\frac{f(x)}{\lambda} \sqrt{15+12 x+10 x^2}+C$ where $C$ is the constant of integration and $f (0)=0$ then $\left( f ^{\prime \prime}(1)+\lambda\right)$ equals

• A. 32
• B. 34
• C. 62
• D. 64

Answer 1

Answer: (A)

$I=\int \frac{\left(x+x^2+x^3\right)}{\sqrt{15+12 x+10 x^2}} d x$
Multiply numerator and denominator by $x ^2$
$\int \frac{x^5+x^4+x^3}{\sqrt{10 x^6+12 x^5+15 x^4}} d x$
Put $10 x ^6+12 x ^5+15 x ^4= t ^2$
$ \Rightarrow 60\left( x ^5+ x ^4+ x ^3\right) dx =2 tdt$
$\frac{1}{30} \int \frac{ t }{ t } dt =\frac{ t }{30}+ C =\frac{ x ^2 \sqrt{10 x ^2+12 x +15}}{30}+ C $
$\therefore f ( x )= x ^2 \text { and } \lambda=30 $
$\Rightarrow f ^{\prime \prime}(1)+ l =32$