Evaluate the integral int frac{1}{1+frac{x}{2}^2}dx

Question
Integrals
asked 2020-11-22
Evaluate the integral \(\int \frac{1}{1+\frac{x}{2}^2}dx\)

Answers (1)

2020-11-23
The first step here will be a substitution to simplify that denominator.
\(u=1+\frac{x}{2}\)
\(\therefore du=\frac{x}{2}dx \rightarrow 2du=dx\)
Substituting these in, we get
\(\int \frac{1}{u^{2}}\times 2du=2\int u^{-2}du\)
Now we just use the power rule for integrals.
\(=2[\frac{u^{-1}}{1}]+C=-\frac{2}{u}+c\)
And finally, we substitute back in for xx and simplify
\(=-\frac{2}{1+\frac{x}{2}+C}=-\frac{4}{2+x}+c\)
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