Evaluate the following integrals:

let $I=\int \frac{1}{x^{2}-10 x+34} d x$

$I=\int \frac{1}{x^{2}-10 x+34} d x$

$=\int \frac{1}{x^{2}+2 x \times 5+(5)^{2}-(5)^{2}+34} d x$

$=\int \frac{1}{(x-5)^{2}-9} d x$

Let $(x-5)=t \ldots \ldots$ (i)

$\Rightarrow d x=d t$

So,

$I=\int \frac{1}{t^{2}+(3)^{2}} d t$

$I=\frac{1}{3} \tan ^{-1}\left(\frac{t}{3}\right)+c$

$\left[\right.$ since, $\left.\int \frac{1}{x^{2}+(a)^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c\right]$

$I=\frac{1}{3} \tan ^{-1}\left(\frac{x-5}{3}\right)+c[u \operatorname{sing}(i)]$

$I=\frac{1}{3} \tan ^{-1}\left(\frac{x-5}{3}\right)+c$