Question

If $\int \frac{x^2-4}{x^4+9x^2+16}dx=A{\tan }^{-1}(f(x))+B$, then the value of $f(10)$ is_____.

Answer 1

Answer: 10.4

$I=\int \frac{x^2-4}{x^4+9x^2+16}dx$
Divide both numerator and denominator by $x^2$
$I=\int \frac{\left(1-\frac{4}{x^2}\right)}{x^2+9+\frac{16}{x^2}}dx=\int \frac{\left(1-\frac{4}{x^2}\right)}{{\left(x+\frac{4}{x}\right)}^2+1}dx$
Put $u=x+\frac{4}{x},du=\left(1-\frac{4}{x^2}\right)dx$
$I=\int \frac{du}{u^2+1}={\tan }^{-1}(u)+B$
$=1\cdot {\tan }^{-1}\left(x+\frac{4}{x}\right)+B$
$\Rightarrow f(x)=x+\frac{4}{x}$
$f(10)=10+\frac{4}{10}=10.4$