Question

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

Answer 1

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