Ikunupe6v

2022-01-05

I am in the middle of a problem and having trouble integrating the following integral:

${\int}_{-1}^{1}\frac{1}{{(1+{x}^{2})}^{2}}dx$

puhnut1m

Beginner2022-01-06Added 33 answers

Try making a substitution $x=\mathrm{tan}u$ . Notice then that

${(1+{x}^{2})}^{2}={(1+{\mathrm{tan}}^{2}u)}^{2}={\left({\mathrm{sec}}^{2}u\right)}^{2}={\left({\mathrm{sec}}^{2}u\right)}^{2}={\mathrm{sec}}^{4}u$

and

$dx={\mathrm{sec}}^{2}udu$

So the indefinite integral is now

$\int \frac{1}{{\mathrm{sec}}^{2}u}du=\int {\mathrm{cos}}^{2}udu$

This new integrand should be easier to integrate. Just remember to change your limits to get the proper evaluation.

and

So the indefinite integral is now

This new integrand should be easier to integrate. Just remember to change your limits to get the proper evaluation.

Robert Pina

Beginner2022-01-07Added 42 answers

If we add and subtract $x}^{2$ in the numerator, we can integrate the first integral immediately

$\int \frac{1}{{(1+{x}^{2})}^{2}}dx=\int \frac{1}{1+{x}^{2}}dx-\int \frac{{x}^{2}}{{(1+{x}^{2})}^{2}}dx$

$\mathrm{arctan}x-\int x\frac{x}{{(1+{x}^{2})}^{2}}dx$

and the second integral by parts:

$\int x\frac{x}{{(1+{x}^{2})}^{2}}dx=x(-\frac{1}{2(1+{x}^{2})})+\int \frac{1}{2(1+{x}^{2})}dx$

$=-\frac{x}{2(1+{x}^{2}\}+\frac{1}{2}\mathrm{arctan}x}$

and the second integral by parts:

user_27qwe

Skilled2022-01-11Added 375 answers

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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