Let $\int\frac{\ln\left(x + \sqrt{1+x^2}\right)}{\sqrt{1+x^2}}dx = fog(x) + c$, where $f(x) = \frac{x^2}{2}$ and $g$ are some functions and $c$ is an arbitrary constant. If $\int f(x)g(x)dx = ax^3g(x) + b\left(1+x^2\right)^{3/2} + c\left(1+x^2\right)^{1/2} + d$, then $\left(\frac{1}{a+b+c}\right)$ is equal to
Step-by-Step Solution
Key Concept: Recognize the integrand as a derivative of a product $f(x)g(x)$ and apply integration by parts with appropriate function identification.
To evaluate $\int \frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}} dx$, recognize that $f(x) = \frac{x^2}{2}$ and $g(x) = \ln(x+\sqrt{1+x^2})$ are related by differentiation. Using the product rule on $f(x)g(x)$ gives $\int f(x)g(x)dx = \frac{x^3}{6}\ln(x+\sqrt{1+x^2}) - \int \frac{x^3}{6\sqrt{1+x^2}}dx$. Substitute $1+x^2 = t^2$ to evaluate the remaining integral, obtaining $\frac{x^3}{6}\ln(x+\sqrt{1+x^2}) - \frac{1}{18}(1+x^2)^{3/2} - \frac{1}{6}(1+x^2)^{1/2} + c$. Computing coefficients: $a = \frac{1}{6}$, $b = -\frac{1}{18}$, $c = -\frac{1}{6}$ yields $a+b+c = \frac{18}{5}$.
Correct Answer: I need to find the value of $\left(\frac{1}{a+b+c}\right)$ based on the given information.
From the step-by-step solution:
- $a = \frac{1}{6}$
- $b = -\frac{1}{18}$
- $c = -\frac{