Question

Let $F\left(x\right)=\int \frac{dx}{(x^2+1)(x^2+9)}$ and $F\left(0\right)=0$, if $F\left(\sqrt{3}\right)=\frac{5}{36}k$, then find the value of $k$.

Answer 1

$F\left(x\right)=\int \frac{dx}{(x^2+1)(x^2+9)}$

$F\left(x\right)=\frac{1}{8}\int \frac{8dx}{(x^2+1)(x^2+9)}$

$F\left(x\right)=\frac{1}{8}\int \left[\frac{(x^2+9)-(x^2+1)}{(x^2+1)(x^2+9)}\right]dx$

$F\left(x\right)=\frac{1}{8}\int [\frac{x^2+9}{(x^2+1)(x^2+9)}-\frac{x^2+1}{(x^2+1)(x^2+9)}]dx$

$F\left(x\right)=\frac{1}{8}\int \frac{dx}{x^2+1}-\int \frac{dx}{x^2+9}$

$F\left(x\right)=\frac{1}{8}[ta{n}^{-1}x-\frac{1}{3}ta{n}^{-1}(\frac{x}{3}\left)\right]+C$

Given $F\left(0\right)=0$

$F\left(0\right)=\frac{1}{8}\left[ta{n}^{-1}\right(0)-\frac{1}{3}ta{n}^{-1}(\frac{0}{3}\left)\right]+C$

$0=0+C$

$\therefore C=0$

$F\left(x\right)=\frac{1}{8}\left[ta{n}^{-1}\right(x)-\frac{1}{3}ta{n}^{-1}(\frac{x}{3}\left)\right]$

$F\left(\sqrt{3}\right)=\frac{1}{8}\left[ta{n}^{-1}\right(\sqrt{3})-\frac{1}{3}ta{n}^{-1}(\frac{\sqrt{3}}{3}\left)\right]$

$F\left(\sqrt{3}\right)=\frac{1}{8}[\frac{\pi }{3}-\frac{\pi }{18}]$

$F\left(\sqrt{3}\right)=\frac{\pi }{24}[\pi -\frac{\pi }{6}]$

$F\left(\sqrt{3}\right)=\frac{5\pi }{144\pi }$

$\frac{5K}{36}=\frac{5\pi }{144\pi }$ $(\because F(\sqrt{3})=\frac{5K}{36}Given)$

$K=\frac{36\pi }{144}$

$K=\frac{\pi }{4}$