Question

If $\int \frac{\left(x+x^2+x^3\right)}{\sqrt{15+12x+10x^2}}dx=\frac{f(x)}{\lambda }\sqrt{15+12x+10x^2}+C$ 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)

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