Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade None
Question:
Let $\int \frac{(f'(x)g(x) - g'(x)f(x))dx}{(f(x) + g(x))\sqrt{f(x)g(x) - g^2(x)}} = \sqrt{m}\tan^{-1}\left(\frac{f(x) - g(x)}{ng(x)}\right) + c$ where $m,n \in \mathbb{N}$ and $C$ is constant of integration ($g(x) > 0$) then the value of $m^2 + n^2$ is
Step-by-Step Solution
Key Concept: Substitution using the ratio $\frac{f(x)}{g(x)}$ converts a complex rational integral into a standard arctangent form.
We start with $I = \int \frac{f'(x)g(x) - g'(x)f(x)}{(f(x) + g(x))\sqrt{f(x)g(x) - g^2(x)}} dx$. Rewriting the denominator using $(g(x))^2$ and setting $\frac{f(x)}{g(x)} - 1 = t^2$, we get $\frac{f'(x)g(x) - g'(x)f(x)}{(g(x))^2} dx = 2tdt$. This transforms the integral to $I = \int \frac{2tdt}{(t^2+2)t} = \frac{2}{\sqrt{2}}\tan^{-1}\left(\frac{t}{\sqrt{2}}\right) + c$. From the final form, comparing coefficients gives $m = 2$ and $n = 2$, so $m^2 + n^2 = 8$.
Correct Answer: 3,4