Tiffany Russell

2021-12-26

Evaluate the integral.
$\int \frac{1}{{x}^{2}+25}dx$

Bob Huerta

Step 1
Consider the given integral:
$\int \frac{1}{{x}^{2}+25}dx$
Step 2
Now, substitute the following:
$x=5\mathrm{tan}\theta$
$dx=\left(5{\mathrm{sec}}^{2}\theta \right)d\theta$
Step 3
Then we will have:
$\int \frac{1}{{x}^{2}+25}dx=\int \frac{1}{25{\mathrm{tan}}^{2}\theta +25}5{\mathrm{sec}}^{2}\theta d\theta$
$=\int \frac{5{\mathrm{sec}}^{2}\theta }{25\left({\mathrm{tan}}^{2}\theta +1\right)}d\theta$
$=\int \frac{5{\mathrm{sec}}^{2}\theta }{25{\mathrm{sec}}^{2}\theta }d\theta$
$=\int \frac{1}{5}d\theta$
$=\frac{\theta }{5}+c$
Step 4
Now, back substitute the value of theta and get the value of required integral:
$\int \frac{1}{{x}^{2}+25}dx=\frac{1}{5}{\mathrm{tan}}^{-1}\left(\frac{x}{5}\right)+c$

ol3i4c5s4hr

Given:
$\int \frac{1}{{x}^{2}+25}dx$
$=\int \frac{5}{25{u}^{2}+25}du$
Simplifying:
$=\frac{1}{5}\int \frac{1}{{u}^{2}+1}du$
$\int \frac{1}{{u}^{2}+1}du$
$=\mathrm{arctan}\left(u\right)$
$\frac{1}{5}\int \frac{1}{{u}^{2}+1}du$
$=\frac{\mathrm{arctan}\left(u\right)}{5}$
$=\frac{\mathrm{arctan}\left(\frac{x}{5}\right)}{5}$
$\int \frac{1}{{x}^{2}+25}dx$
$=\frac{\mathrm{arctan}\left(\frac{x}{5}\right)}{5}+C$

karton

Given:
$\int \frac{1}{{x}^{2}+25}dx$
Evaluate the integral
$\frac{1}{5}×\mathrm{arctan}\left(\frac{x}{5}\right)$
Calculate
$\frac{\mathrm{arctan}\left(\frac{x}{5}\right)}{5}$
$\frac{\mathrm{arctan}\left(\frac{x}{5}\right)}{5}+C$